Proof of Theorem 4. We need to show that, under the assumption of \(\mathsf{T}\) being unique and maximal, if \(g(t)\) is right \(\mathsf{T}\)-invariant, then \(\mathsf{T}\) is also rigid (see Definition 2). Let \(\mathfrak{t} = {\rm Lie}(\mathsf{T})\) and consider the \(Q\)-orthogonal splitting \[\mathfrak{m} = \mathfrak{t} + \mathfrak{t}^{\perp} \,\, .\] By maximality of \(\mathfrak{t}\), there exists no trivial \({\rm Ad}(\mathsf{H}\mathsf{T})\)-submodule in \(\mathfrak{t}^{\perp}\). Therefore, since \(g(t)\) is right \(\mathsf{T}\)-invariant for any \(t < 0\), it follows from Schur’s Lemma that \(\mathfrak{t}\) and its \(Q\)-orthogonal complement \(\mathfrak{t}^{\perp}\) are \(g(t)\)-orthogonal for any \(t < 0\). By maximality of \(\mathfrak{t}\) and Theorem 1, there exists \(\delta>0\) such that \(|X|_{g(t)} \geq \delta |X|_{g(-1)}\) for any \(X \in \mathfrak{t}^{\perp}\) and for any \(t \leq -1\). Finally, since the splitting \(\mathfrak{m} = \mathfrak{t} + \mathfrak{t}^{\perp}\) is \(g(t)\)-orthogonal for any \(t < 0\) and the collapsing torus \(\mathsf{T}\) is unique, it follows that \(\lim_{t \to -\infty}|t|^{-\frac{1}{2}}|V|_{g(t)} = 0\) for any \(V \in \mathfrak{t}\), because every sequence of times \(t^{(n)} \to -\infty\) admits a subsequence along which the limit is zero (see the first equation in 14 ). ◻

Proof. We will give the proof of the inequalities for \(|\hat{A}_g|_g\), since the proof for \(|\check{A}_g|_g\) is very similar. For convenience, we set \(A_{pq}^r \mathrel{\vcenter{:}}= g(A_{X_p}X_q, U_r)\). Since \[\begin{align} |A|_g^2 &= \sum_{p,q,r}g(A_{X_p}X_q, U_r)^2 = \sum_{p,q,r} (A_{pq}^r)^2 \,\, , \\ |\hat{A}_g|_g^2 &= \sum_{i,j} \hat{A}_g(U_i, U_j)^2 = \sum_{i,j} \bigg( \sum_{k,l} A_{kl}^i A_{kl}^j \bigg)^2 \,\, , \end{align}\] the Cauchy–Schwarz inequality implies that \[|\hat{A}_g|_g^2 \leq \sum_{i,j} \bigg(\bigg(\sum_{k,l}(A_{kl}^i)^2 \bigg) \bigg(\sum_{k,l}(A_{kl}^j)^2 \bigg)\bigg) \leq m^2 |A|_g^4 \,\, .\] On the other hand, by the equivalence of norms on a finite-dimensional space, we obtain \[|\hat{A}_g|_g^2 \geq C_1(m) \bigg(\sum_{i,j} \bigg| \sum_{k,l}A_{kl}^iA_{kl}^j \bigg| \bigg)^2 \geq C_1(m) \bigg( \sum_{i,k,l} \big|A_{kl}^i\big| \cdot \big|A_{kl}^i\big| \bigg)^2 = C_1(m) |A|_g^4\] and this concludes the proof. ◻

Proof. Let \(\{X_i\}\) be a \(g(1)\)-orthonormal basis for \(\mathfrak{b}\) and denote by \(X_i(\tau)\) the vector fields evolving through \[\frac{\rm d}{{\rm d}\tau}X_i(\tau) = -{\rm Ric}(g(\tau))^{\sharp}(X_i(\tau)) \,\, , \quad X_i(1) = X_i \,\, .\] Then, \(\{X_i(\tau)\}\) is \(g(\tau)\)-orthonormal. Notice that \(A\) itself does not depend on \(g\), so that \[\begin{align} \frac{\rm d}{{\rm d}\tau}|A|_{g(\tau)}^2 &= \frac{\rm d}{{\rm d}\tau} \left\{\sum_{i,j} g(A_{X_i}X_j,A_{X_i}X_j)\right\} \\ &= \sum_{i,j} 2{\rm Ric}(g(\tau))(A_{X_i}X_j,A_{X_i}X_j) -2g(A_{({\rm Ric}(g(\tau))^{\sharp}X_i)}X_j,A_{X_i}X_j) \\ &\qquad\qquad -2g(A_{X_i}({\rm Ric}(g(\tau))^{\sharp}X_j),A_{X_i}X_j) \\ &= \sum_{i,j} 2{\rm Ric}(g(\tau))(A_{X_i}X_j,A_{X_i}X_j) -4g(A_{X_i}({\rm Ric}(g(\tau))^{\sharp}X_j),A_{X_i}X_j) \,\, . \end{align}\] Notice that the fibers of \(M \to B\) are abelian and totally geodesic. At any fixed time \(\tau > 0\), we can consider a \(g(\tau)\)-orthonormal basis \(\{U_p\}\) for \(\mathfrak{t}\). Therefore, by [45], we have \[\begin{align} 2\sum_{i,j} {\rm Ric}(g(\tau))(A_{X_i}X_j,A_{X_i}X_j) &= 2\sum_{i,j} \hat{A}_{g(\tau)}(A_{X_i}X_j,A_{X_i}X_j) \\ &= 2\sum_{i,j,p,q} \hat{A}_{g(\tau)}(U_p,U_q)g(\tau)(A_{X_i}U_p,X_j)g(\tau)(A_{X_i}U_q,X_j) \\ &= 2\sum_{i,j,p,q} \hat{A}_{g(\tau)}(U_p,U_q)^2 \\ &= 2|\hat{A}_{g(\tau)}|_{g(\tau)}^2 \end{align}\] and \[\begin{align} -4\sum_{i,j}g(A_{X_i}({\rm Ric}(g(\tau))^{\sharp}X_j),A_{X_i}X_j) &= 4\sum_{i,j}{\rm Ric}(g(\tau))(A_{X_i}A_{X_i}X_j,X_j) \\ &= -4\sum_{i,j,k}{\rm Ric}(g(\tau))(X_j,X_k)g(\tau)(A_{X_i}X_k,A_{X_i}X_j) \\ &= -4\sum_{i,j,k}{\rm Ric}(g(\tau))(X_j,X_k)\check{A}_{g(\tau)}(X_j,X_k) \\ &= -4g(\tau)({\rm Ric}(g(\tau)),\check{A}_{g(\tau)}) \,\, . \end{align}\] Finally, by 22 we obtain 28 . ◻

Proof. Since the collapsing torus \(\mathsf{T}\) is rigid, it follows that \[\begin{gather} \frac{x_k(\tau)}{\tau} \to 0 \quad \text{as \tau \to +\infty, for every 1 \leq k \leq s} \,\, , \\ \frac{x_i(\tau)}{\tau} \geq \delta \quad \text{for any \tau \geq 1, for every s < i \leq \ell} \,\, . \end{gather}\] Therefore, by 21 we obtain \[\lim_{\tau \to +\infty} |A|_{\frac{1}{\tau}g(\tau)} = 0 \,\, ,\] which completes the proof. ◻

Proof of Theorem 3. By [45] and homogeneity, we have \[\frac{\rm d}{{\rm d}\tau} {\rm scal}(g(\tau)) = -2|{\rm Ric}(g)|_g^2\] and, by [45], \[\frac{\rm d}{{\rm d}\tau} {\rm vol}_B(g_{\mathfrak{b}}(\tau)) = {\rm Tr}({\rm Ric}(g)|_{\mathfrak{b}}){\rm vol}_B(g_{\mathfrak{b}}) = ({\rm scal}_B(g_{\mathfrak{b}})-2|A|_g^2){\rm vol}_B(g_{\mathfrak{b}}) \,\, .\] Therefore, by using 28 , we have \[\begin{align} \frac{\rm d}{{\rm d}\tau}F(g(\tau)) &= 2\Big(-|{\rm Ric}(g)|_g^2 +2|\hat{A}_g|_g^2 +8|\check{A}_g|_g^2 -4g({\rm Ric}_B(g_{\mathfrak{b}}),\check{A}_g)\Big){\rm vol}_B(g_{\mathfrak{b}})^{\frac{2}{k}} +\\ &\quad\quad +\tfrac2k\big({\rm scal}(g)+2|A|_{g}^2\big)({\rm scal}_B(g_{\mathfrak{b}})-2|A|_g^2) {\rm vol}_B(g_{\mathfrak{b}})^{\frac{2}{k}} \\ &= -2{\rm vol}_B(g_{\mathfrak{b}})^{\frac{2}{k}} \Big(|{\rm Ric}(g)|_g^2 -|\hat{A}_g|_g^2 -4|\check{A}_g|_g^2 +4g({\rm Ric}_B(g_{\mathfrak{b}}),\check{A}_g) -\tfrac1k{\rm scal}_B(g_{\mathfrak{b}})^2 \\ &\quad\quad +\tfrac1k{\rm scal}_B(g_{\mathfrak{b}})^2|A|_g^2 -|\hat{A}_g|_g^2 -4|\check{A}_g|_g^2 +\tfrac2k|A|_g^4\Big) \,\, . \end{align}\] Since \[|{\rm Ric}(g)|_g^2 = |\hat{A}_g|_g^2 +|{\rm Ric}_B(g_{\mathfrak{b}})|_{g_{\mathfrak{b}}}^2 +4|\check{A}_g|_g^2 -4g({\rm Ric}_B(g_{\mathfrak{b}}),\check{A}_g) \,\, ,\] we obtain 25 . Moreover, by the Cauchy–Schwarz inequality, we have \[|{\rm Ric}_B(g_{\mathfrak{b}}(\tau))|_{g_{\mathfrak{b}}(\tau)}^2-\tfrac1k{\rm scal}_B(g_{\mathfrak{b}}(\tau))^2 \geq 0\] with equality if and only if \({\rm Ric}^0_B(g_{\mathfrak{b}}(\tau)) = 0\). By ?? , the remaining term can be estimated as \[\tfrac1k{\rm scal}_B(g_{\mathfrak{b}}(\tau))|A|_{g(\tau)}^2 -|\hat{A}_{g(\tau)}|_{g(\tau)}^2 -4|\check{A}_{g(\tau)}|_{g(\tau)}^2 +\tfrac2k|A|_{g(\tau)}^4 \geq \tfrac1{k}|A|_{g(\tau)}^2\big({\rm scal}_B(g_{\mathfrak{b}}(\tau)) -C|A|_{g(\tau)}^2\big)\] for some \(C>0\). Therefore, by 29 , there exists \(T>0\) such that \[{\rm scal}_B(g_{\mathfrak{b}}(\tau)) \geq C|A|_{g(\tau)}^2 \quad \text{for any \tau \geq T}\] and this completes the proof of 26 . ◻

Proof of Theorem 2. Let \(g(\tau)\) be a collapsed solution to the backward Ricci flow existing for all \(\tau > 0\) and \(\bar{g}(\tau) = \frac{1}{\tau}g(\tau)\) the corresponding curvature-normalized metrics. Assume that the collapsing torus \(\mathsf{T}\) of \(g(\tau)\) is rigid, and that \(g(\tau)\) is right \(\mathsf{T}\)-invariant. Then, \(\bar{g}(\tau)\) splits as in 27 , with \[\bar{g}_{\mathfrak{t}}(\tau) \to 0 \,\, \text{ as \tau \to +\infty} \,\, , \qquad \bar{g}_{\mathfrak{b}}(\tau) \geq \delta\, \bar{g}_{\mathfrak{b}}(1) \,\, \text{ for any \tau \geq 1}.\] By 29 and [45], \({\rm scal}(\bar{g}_{\mathfrak{b}}(\tau))\) converges to a positive constant as \(\tau \to +\infty\). Since \(F(\bar{g}(\tau))\) is bounded, it follows that \({\rm vol}_B(\bar{g}_{\mathfrak{b}}(\tau))\) is bounded as well. Therefore, the 1-parameter family \(\bar{g}_{\mathfrak{b}}(\tau)\) lives in a compact set of \(\mathsf{G}\)-invariant metrics on \(B\). We now prove the following two claims.

To prove Claim 1, we observe that, by Theorem 3, \[\begin{gather} \tau\frac{\rm d}{{\rm d}\tau}F(g(\tau)) = -2{\rm vol}_B(\bar{g}_{\mathfrak{b}})^{\frac{2}{k}} \tau{\rm scal}(g) \Big( |{\rm Ric}_B(\bar{g}_{\mathfrak{b}})|_{\bar{g}_{\mathfrak{b}}}^2 -\tfrac1k{\rm scal}_B(\bar{g}_{\mathfrak{b}})^2 \\ +\tfrac1k{\rm scal}_B(\bar{g}_{\mathfrak{b}})|A|_{\bar{g}_{\mathfrak{b}}}^2 -|\hat{A}_{\bar{g}}|_{\bar{g}}^2 -4|\check{A}_{\bar{g}}|_{\bar{g}}^2 +\tfrac2k|A|_{\bar{g}}^4\Big) \,\, . \end{gather}\] We claim that there exists \(\tau_i \to +\infty\) such that \(\tau_i\frac{\rm d}{{\rm d}\tau}F(g(\tau_i)) \to 0\). Indeed, suppose that this is not the case. Then, by 26 , there exist \(T>0\) and \(C>0\) such that \[\label{eq:absurdF39} \tau \frac{\rm d}{{\rm d}\tau}F(g(\tau)) \leq -C \quad \text{ for any \tau \geq T} \,\, .\tag{30}\] Then, define the function \[\psi: [T,+\infty) \to \mathbb{R} \,\, , \quad \psi(\tau) \mathrel{\vcenter{:}}= F(g(T)) -C\log\left(\frac{\tau}{T}\right)\] and notice that \(\psi(T) = F(g(T))\), and \(\lim_{\tau \to +\infty} \psi(\tau) = - \infty\). Moreover, by 30 , it follows that \(\frac{\rm d}{{\rm d}\tau}F(g(\tau)) \leq \psi'(\tau)\) for all \(\tau \geq T\), from which we obtain \(\lim_{\tau \to +\infty} F(g(\tau)) = -\infty\). This is a contradiction since \(F(g(\tau))\) is non-negative.

Consider \(\{\tau_i\}\) as above. Since the quantities \({\rm vol}_B(\bar{g}_{\mathfrak{b}})^{\frac{2}{k}}\), \(\tau{\rm scal}(g)\), and \({\rm scal}_B(\bar{g}_{\mathfrak{b}})\) are bounded from above and bounded away from zero along the flow, and \(|A|_{\bar{g}}\to 0\) by 29 , it follows that \[|{\rm Ric}_B(\bar{g}_{\mathfrak{b}}(\tau_i))|_{\bar{g}_{\mathfrak{b}}(\tau_i)}^2-\tfrac1k{\rm scal}_B(\bar{g}_{\mathfrak{b}}(\tau_i))^2 \to 0 \,\, .\] Up to passing to a subsequence, we can assume that \(\bar{g}_{\mathfrak{b}}(\tau_i) \to \bar{g}_{\mathfrak{b}}^{\infty}\) as \(i \to +\infty\). Then, Claim 1 follows by applying the Cauchy–Schwarz inequality.

By the previous step, we have convergence to the degenerate metric \(0 \oplus \bar{g}_{\mathfrak{b}}^{\infty}\) along a sequence of times. Here, we need to show full convergence along the flow. We use the theory the second- and third-named authors developed in [34] to prove this fact.

By [34], the Ricci curvature can be analytically extended to the space \[S^2(\mathfrak{t}^*) \oplus S^2_+(\mathfrak{b}^*)^{{\rm Ad}(\mathsf{H}\mathsf{T})}\] of submersion metrics on \(M = \mathsf{G}/\mathsf{H}\) that are possibly not positive-definite on the toral fibers of 18 . Moreover, by [45], the scalar curvature of a submersion metric \(g\) is given by \[{\rm scal}(g) = {\rm scal}_B(g_{\mathfrak{b}}) -|A|_g^2\] and so, by 21 , it follows that the scalar curvature can be analytically extended to the space \(S^2(\mathfrak{t}^*) \oplus S^2_+(\mathfrak{b}^*)^{{\rm Ad}(\mathsf{H}\mathsf{T})}\) as well. By [34], there exists an inner product \(\langle\!\langle \cdot, \cdot \rangle\!\rangle^{\bar{g}_{\mathfrak{b}}^{\infty}}\) on \(S^2(\mathfrak{m}^*)^{{\rm Ad}(\mathsf{H})}\), depending on the Einstein metric \(\bar{g}_{\mathfrak{b}}^{\infty}\) on the base, such that the following claims hold. First, up to time reparametrization, the projection of \(g(\tau)\) on the unit sphere \[\Sigma \mathrel{\vcenter{:}}= \big\{g \in S^2(\mathfrak{t}^*) \oplus S^2_+(\mathfrak{b}^*)^{{\rm Ad}(\mathsf{H}\mathsf{T})} : \langle\!\langle g, g \rangle\!\rangle^{\bar{g}_{\mathfrak{b}}^{\infty}} = 1\big\}\] is a solution to the gradient flow of the restricted scalar curvature functional \[\label{eq:restrscal} g \in \Sigma \mapsto {\rm scal}(g) \,\, .\tag{31}\] Second, the degenerate metric \(0 \oplus \bar{g}_{\mathfrak{b}}^{\infty}\) is a critical point for the functional 31 . Therefore, by [50], the convergence is uniform along the flow trajectories, and this concludes the proof of Claim 2.

Finally, arguing as in [34], it follows that \((\mathsf{G}/\mathsf{H},\bar{g}(\tau))\) converges to \((\mathsf{G}/\mathsf{H}\mathsf{T},\bar{g}_{\mathfrak{b}}^{\infty})\) in the Gromov-Hausdorff topology as \(\tau \to +\infty\). ◻

Proof. Fix \(1 \leq p \leq \ell_0\), \(1 \leq i \leq \ell\) and assume that there exists \(1 \leq j \leq \ell\) such that \([pij] >0\). By means of Remark 1, the image \(\widetilde{\mathfrak{m}}_i \mathrel{\vcenter{:}}= {\rm ad}(V_p)(\mathfrak{m}_i)\) is \({\rm Ad}(\mathsf{H})\)-equivalent to \(\mathfrak{m}_i\) and thus it is contained in the linear span of the isotypical class of \(\mathfrak{m}_i\). By assumption, there exists \(1 \leq j' \leq \ell\) such that \(\widetilde{\mathfrak{m}}_i \cap \mathfrak{m}_{j'} \neq \{0\}\) and hence \(\widetilde{\mathfrak{m}}_i = \mathfrak{m}_{j'}\). By hypothesis, \(j' = j\) and the claim follows. ◻

Proof. It is sufficient to prove the lemma for \(i \in I_p\). Equation 37 is true since \(\mathfrak{m}_i\) and \(\mathfrak{m}_{\phi_p(i)}\) are \({\rm Ad}(\mathsf{H})\)-equivalent and so, in particular, are isomorphic as linear spaces. Moreover, since \(\mathfrak{m}_i\) and \(\mathfrak{m}_{\phi_p(i)}\) are intertwined by the adjoint action of a vector \(V_p \in \mathfrak{m}_0\), it follows that both \(\mathfrak{m}_i\) and \(\mathfrak{m}_{\phi_p(i)}\) are contained in a single simple ideal of \(\mathfrak{g}\). Therefore, since \(Q\) is \({\rm Ad}(\mathsf{G})\)-invariant, 38 follows. ◻

Proof. Fix \(1 \leq p \leq \ell_0\), \(i \in I_p\) and, for the sake of shortness, set \(\overline{r} \mathrel{\vcenter{:}}= \phi_p(r)\) for any \(1 \leq r \leq \ell\). Fix a \(\varphi\)-adapted, \(Q\)-orthonormal basis \(\EuScript{B}\) for \(\mathfrak{m}\) and assume that it diagonalizes \({\rm ad}(V_p)\) as in 34 . Then, for any \(1 \leq j, k \leq \ell\) we have \[\label{eq:Jacobi} \begin{gather} \mu_{p,k} Q([e_{\bar{\alpha}},e_{\beta}],e_{\bar{\gamma}}) +\mu_{p,j} Q([e_{\bar{\alpha}},e_{\bar{\beta}}],e_{\gamma}) -\mu_{p,i}Q([e_{\alpha},e_{\beta}],e_{\gamma}) = 0 \,\, , \\ \mu_{p,k}Q([e_{\alpha},e_{\bar{\beta}}],e_{\bar{\gamma}}) +\mu_{p,i} Q([e_{\bar{\alpha}},e_{\bar{\beta}}],e_{\gamma}) -\mu_{p,j}Q([e_{\alpha},e_{\beta}],e_{\gamma}) = 0 \end{gather}\tag{42}\] for any \(e_{\alpha} \in \EuScript{B} \cap \mathfrak{m}_i\), \(e_{\beta} \in \EuScript{B} \cap \mathfrak{m}_j\), \(e_{\gamma} \in \EuScript{B} \cap \mathfrak{m}_k\). Indeed, by 34 , the \({\rm Ad}(\mathsf{G})\)-invariance of \(Q\) and the Jacobi Identity, it follows that \[\begin{gather} \mu_{p,k} Q([e_{\bar{\alpha}},e_{\beta}],e_{\bar{\gamma}}) +\mu_{p,j} Q([e_{\bar{\alpha}},e_{\bar{\beta}}],e_{\gamma}) -\mu_{p,i}Q([e_{\alpha},e_{\beta}],e_{\gamma}) = \\ = -Q([V_p,[e_{\bar{\alpha}},e_{\beta}]],e_{\gamma}) -Q([e_{\bar{\alpha}},[e_{\beta},V_p]],e_{\gamma}) -Q([e_{\beta},[V_p,e_{\bar{\alpha}}]],e_{\gamma}) = 0 \,\, . \end{gather}\] The second equation in 42 is obtained similarly.

By taking the squares of 42 , we have \[\begin{align} \mu_{p,k}^2 Q([e_{\bar{\alpha}},e_{\beta}],e_{\bar{\gamma}})^2 &= \mu_{p,j}^2 Q([e_{\bar{\alpha}},e_{\bar{\beta}}],e_{\gamma})^2 +\mu_{p,i}^2Q([e_{\alpha},e_{\beta}],e_{\gamma})^2 \\ &\qquad\qquad\qquad -2\mu_{p,i}\mu_{p,j} Q([e_{\bar{\alpha}},e_{\bar{\beta}}],e_{\gamma}) Q([e_{\alpha},e_{\beta}],e_{\gamma}) \,\, ,\\ \mu_{p,k}^2 Q([e_{\alpha},e_{\bar{\beta}}],e_{\bar{\gamma}}))^2 &= \mu_{p,i}^2 Q([e_{\bar{\alpha}},e_{\bar{\beta}}],e_{\gamma})^2 +\mu_{p,j}^2Q([e_{\alpha},e_{\beta}],e_{\gamma})^2 \\ &\qquad\qquad\qquad -2\mu_{p,i} \mu_{p,j}Q([e_{\bar{\alpha}},e_{\bar{\beta}}],e_{\gamma})Q([e_{\alpha},e_{\beta}],e_{\gamma}), \end{align}\] and so \[\begin{gather} \label{eq:Jacobi2} \mu_{p,k}^2 Q([e_{\bar{\alpha}},e_{\beta}],e_{\bar{\gamma}})^2 -\mu_{p,j}^2 Q([e_{\bar{\alpha}},e_{\bar{\beta}}],e_{\gamma})^2 -\mu_{p,i}^2Q([e_{\alpha},e_{\beta}],e_{\gamma})^2 = \\ \mu_{p,k}^2 Q([e_{\alpha},e_{\bar{\beta}}],e_{\bar{\gamma}}))^2 -\mu_{p,i}^2 Q([e_{\bar{\alpha}},e_{\bar{\beta}}],e_{\gamma})^2 -\mu_{p,j}^2Q([e_{\alpha},e_{\beta}],e_{\gamma})^2 \,\, . \end{gather}\tag{43}\] Taking sums in 43 , we get \[\label{eq:proofcoef} \mu_{p,k}^2 [\bar{i}j\bar{k}] -\mu_{p,j}^2 [\bar{i}\bar{j}k] -\mu_{p,i}^2 [ijk] = \mu_{p,k}^2 [i\bar{j}\bar{k}] -\mu_{p,i}^2 [\bar{i}\bar{j}k] -\mu_{p,j}^2 [ijk] \,\, .\tag{44}\] By applying cyclic index permutations and the index transformations \[(j,k) \mapsto (\bar{j},\bar{k}) \,\, , \quad (i,j) \mapsto (\bar{i},\bar{j}) \,\, , \quad (i,k) \mapsto (\bar{i},\bar{k})\] to 44 , we obtain the following linear system: \[\label{eq:linsyst1} \left\{\!\!\begin{array}{rrrrr} \mu_{p,i}^2 [\bar{i}\bar{j}k] &\!\!\!\! +(\mu_{p,j}^2 -\mu_{p,k}^2) [i\bar{j}\bar{k}] &\!\!\!\! -\mu_{p,i}^2 [\bar{i}j\bar{k}] &\!\!\!\! -(\mu_{p,j}^2 -\mu_{p,k}^2) [ijk] &\!\!\!\! = 0 \\ \mu_{p,j}^2 [\bar{i}\bar{j}k] &\!\!\!\! -\mu_{p,j}^2 [i\bar{j}\bar{k}] &\!\!\!\! +(\mu_{p,i}^2 -\mu_{p,k}^2) [\bar{i}j\bar{k}] &\!\!\!\! -(\mu_{p,i}^2 -\mu_{p,k}^2) [ijk] &\!\!\!\! = 0 \\ \mu_{p,k}^2 [\bar{i}\bar{j}k] &\!\!\!\! -(\mu_{p,i}^2 -\mu_{p,j}^2) [i\bar{j}\bar{k}] &\!\!\!\! +(\mu_{p,i}^2 -\mu_{p,j}^2) [\bar{i}j\bar{k}] &\!\!\!\! -\mu_{p,k}^2 [ijk] &\!\!\!\! = 0 \\ (\mu_{p,i}^2 -\mu_{p,j}^2) [\bar{i}\bar{j}k] &\!\!\!\! -\mu_{p,k}^2 [i\bar{j}\bar{k}] &\!\!\!\! +\mu_{p,k}^2 [\bar{i}j\bar{k}] &\!\!\!\! -(\mu_{p,i}^2 -\mu_{p,j}^2) [ijk] &\!\!\!\! = 0 \\ (\mu_{p,j}^2 -\mu_{p,k}^2) [\bar{i}\bar{j}k] &\!\!\!\! +\mu_{p,i}^2 [i\bar{j}\bar{k}] &\!\!\!\! -(\mu_{p,j}^2 -\mu_{p,k}^2) [\bar{i}j\bar{k}] &\!\!\!\! -\mu_{p,i}^2 [ijk] &\!\!\!\! = 0 \\ (\mu_{p,i}^2 -\mu_{p,k}^2) [\bar{i}\bar{j}k] &\!\!\!\! -(\mu_{p,i}^2 -\mu_{p,k}^2) [i\bar{j}\bar{k}] &\!\!\!\! +\mu_{p,j}^2 [\bar{i}j\bar{k}] &\!\!\!\! -\mu_{p,j}^2 [ijk] &\!\!\!\! = 0 \end{array} \right. \,\, .\tag{45}\] Let us write the the system 45 in the form \[A \cdot \big([\bar{i}\bar{j}k], [i\bar{j}\bar{k}], [\bar{i}j\bar{k}], [ijk]\big)^t = 0 \,\, .\] Then, the matrix \(A\) factorizes as the product \(A = P\cdot A'\), where \(P\) is the invertible matrix \[P \mathrel{\vcenter{:}}= \left(\!\!\begin{array}{cccccc} \tfrac12 & -\tfrac12 & \tfrac12 & \tfrac12 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ 1 & 1 & 0 & 0 & 0 & -1 \\ -\tfrac12 & \tfrac12 & \tfrac12 & \tfrac12 & 0 & 0 \\ 0 & 0 & 1 & -1 & -1 & 0 \end{array}\!\!\right)\] and \(A'\) is given by \[A' \mathrel{\vcenter{:}}= \left(\!\!\begin{array}{cccc} \mu_{p,i}^2 -\mu_{p,j}^2 +\mu_{p,k}^2 & -\mu_{p,i}^2 +\mu_{p,j}^2 -\mu_{p,k}^2 & 0 & 0 \\[2pt] 0 & 0 & \mu_{p,i}^2 -\mu_{p,j}^2 +\mu_{p,k}^2 & -\mu_{p,i}^2 +\mu_{p,j}^2 -\mu_{p,k}^2 \\[2pt] \mu_{p,i}^2 +\mu_{p,j}^2 -\mu_{p,k}^2 & 0 & 0 & -\mu_{p,i}^2 -\mu_{p,j}^2 +\mu_{p,k}^2 \\[2pt] 0 & \mu_{p,i}^2 +\mu_{p,j}^2 -\mu_{p,k}^2 & -\mu_{p,i}^2 -\mu_{p,j}^2 +\mu_{p,k}^2 & 0 \\[2pt] \mu_{p,j}^2 & -\mu_{p,j}^2 & \mu_{p,i}^2 -\mu_{p,k}^2 & -\mu_{p,i}^2 +\mu_{p,k}^2 \\[2pt] \mu_{p,k}^2 & -\mu_{p,i}^2 +\mu_{p,j}^2 & \mu_{p,i}^2 -\mu_{p,j}^2 & -\mu_{p,k}^2 \end{array}\!\!\right) \,\, .\]

Notice that, since \(\mu_{p,i} \neq 0\) by assumption, the quantities \[\mu_{p,i}^2 -\mu_{p,j}^2 +\mu_{p,k}^2 \,\, , \quad \mu_{p,i}^2 +\mu_{p,j}^2 -\mu_{p,k}^2\] cannot vanish simultaneously. Then, by explicit computation, the row-echelon form of \(A'\) is as described in the following table.

Therefore, we obtain the following characterization:

• if \(\mu_{p,j} = 0\) and \(\mu_{p,k}^2 = \mu_{p,i}^2\), then \({\rm rank}(A) = 2\) and the solutions to 45 are \[[\bar{i}\bar{j}k] = [i\bar{j}\bar{k}] \,\, , \quad [\bar{i}j\bar{k}] = [ijk] \,\, ;\]

• if \(\mu_{p,k} = 0\) and \(\mu_{p,j}^2 = \mu_{p,i}^2\), then \({\rm rank}(A) = 2\) and the solutions to 45 are \[[\bar{i}\bar{j}k] = [ijk] \,\, , \quad [i\bar{j}\bar{k}] = [\bar{i}j\bar{k}] \,\, ;\]

• in all the other cases, \({\rm rank}(A) = 3\) and the solutions to 45 are \[[\bar{i}\bar{j}k] = [i\bar{j}\bar{k}] = [\bar{i}j\bar{k}] = [ijk] \,\, .\]

Based on this characterization, we proceed as follows.

• If \(j \in I^+_p\), then \(\mu_{p,j}^2 > 0\), and we are either in case \((2)\) or \((3)\). Hence \([\bar{i}\bar{j}k] = [ijk]\), that is 39 .

• If \(j,k \in I^0_p\), then \(\mu_{p,j} = \mu_{p,k} = 0\), so we are in case \((3)\). Moreover, since \(j=\bar{j}\) and \(k=\bar{k}\), we obtain \([\bar{i}jk] = [ijk]\), that is 40 .

• Let \(1 \leq q \leq \ell_0\) be such that \([V_p,V_q]=0\) and set \(j \mathrel{\vcenter{:}}= \phi_q(i)\), \(k \mathrel{\vcenter{:}}= q\). Then \(\mu_{p,k} = 0\), and we are again in case \((2)\) or \((3)\). Hence \([\bar{i}\bar{j}k] = [ijk]\). By 36 , this gives 41 .

This concludes the proof. ◻

Proof. Let \(\varphi\) be stably Ricci-diagonal and consider the subset \[D \mathrel{\vcenter{:}}= \{t \in I : \text{g(t) is \varphi-diagonal} \} \,\, .\] Then, \(D\) is non-empty and closed. We show below that \(D\) is also open, which leads to \(D=I\).

Fix \(t_1 \in D\) and consider the following system of ODEs: \[\label{eq:diagsyst} \dot{x}_i(t) = -2{\rm ric}_i\big(x_1(t),{\dots},x_{\ell}(t)\big) \,\, , \quad 1 \leq i \leq \ell\tag{47}\] where the functions \({\rm ric}_i\) are defined in 12 . By the existence theorem for ODEs, there exists a solution \(\big(x_1(t),{\dots},x_{\ell}(t)\big)\) to 47 defined for \(t \in (t_1-\epsilon,t_1+\epsilon)\), for some \(\epsilon >0\), such that \(\big(x_1(t_1),{\dots},x_{\ell}(t_1)\big)\) is the tuple of eigenvalues of \(g(t_1)\). Since \(g(t_1)\) is positive definite, up to choosing a smaller \(\epsilon\), it follows that \[h(t) \mathrel{\vcenter{:}}= x_1(t)\,Q|_{\mathfrak{m}_1 \otimes \mathfrak{m}_1} + {\dots} + x_{\ell}(t)\,Q|_{\mathfrak{m}_{\ell} \otimes \mathfrak{m}_{\ell}}\] is a \(\mathsf{G}\)-invariant Riemannian metric on \(M\) and \((t_1-\epsilon,t_1+\epsilon) \subset I\). Since \(\varphi\) is stably Ricci-diagonal, it follows that \(h(t)\) is a Ricci flow solution. Therefore, by uniqueness of solutions to the Ricci flow on compact manifolds, \(h(t) = g(t)\) for every \(t \in (t_1-\epsilon,t_1+\epsilon)\). In particular, \((t_1-\epsilon,t_1+\epsilon) \subset D\). ◻

Proof of Proposition 3. By hypotheses and Proposition 3 the metric \(g(t)\) is diagonal, i.e., we can write \[g(t) = x_1(t)\,Q|_{\mathfrak{m}_1 \otimes \mathfrak{m}_1} + {\dots} + x_{\ell}(t)\,Q|_{\mathfrak{m}_{\ell} \otimes \mathfrak{m}_{\ell}}\] for some NR-decomposition, where \(x_i: (-\infty,0) \to (0,+\infty)\) are smooth functions. Since \(N_{\mathsf{G}}(\mathsf{H})/\mathsf{H}\) has rank \(1\), and its Lie algebra is identified with \(\mathfrak{m}_0\) (see Section 2.2), either \(\mathfrak{m}_0 = \mathfrak{m}_1 \simeq \mathbb{R}\), or \(\mathfrak{m}_0 = \mathfrak{m}_1 +\mathfrak{m}_2 +\mathfrak{m}_3 \simeq \mathfrak{so}(3)\). If \(\mathfrak{m}_0 = \mathfrak{m}_1 \simeq \mathbb{R}\), then the collapsing torus is necessarily unique, and its Lie algebra coincides with all of \(\mathfrak{m}_0\).

Assume now that \(\mathfrak{m}_0 = \mathfrak{m}_1 +\mathfrak{m}_2 +\mathfrak{m}_3 \simeq \mathfrak{so}(3)\) and consider the continuous function \[\lambda(t) \mathrel{\vcenter{:}}= \min\{x_1(t), x_2(t), x_3(t)\} \,\, .\] Notice that \(\tfrac{\lambda(t)}{|t|} \to 0\) as \(t \to -\infty\). Indeed, using Theorem 1, we see that for any sequence of times \(t^{(n)} \to -\infty\), there exist a subsequence \(\{t^{(n_k)}\} \subset \{t^{(n)}\}\) such that \[\frac{\lambda(t^{(n_k)})}{|t^{(n_k)}|} \to 0\] as \(k \to +\infty\). Consider also the set of all times where the minimum value \(\lambda(t)\) is achieved by more than one eigenvalues, that is \[\EuScript{X} \mathrel{\vcenter{:}}= \big\{t \in (-\infty,-1] : \text{ there exist 1 \leq i < j \leq 3 such that \lambda(t) = x_i(t) = x_j(t)}\big\} \,\, .\] We claim that there exists \(T>1\) such that \(\EuScript{X} \subset [-T,-1]\), i.e., there exists an eigenvalue which is strictly less than the other two for all times \(t < -T\).

In fact, assume by contradiction that there is a sequence of times \(\{t^{(n)}\} \subset \EuScript{X}\) such that \(t^{(n)} \to -\infty\). Then, up to passing to a subsequence, there exist \(1 \leq i < j \leq 3\) such that \[\lambda(t^{(n)}) = x_i(t^{(n)}) = x_j(t^{(n)}) \quad \text{ for any n \in \mathbb{N}} \,\, ,\] which in turn implies that \[\lim_{n \to +\infty}\frac{x_i(t^{(n)})}{|t^{(n)}|} = \lim_{n \to +\infty}\frac{x_j(t^{(n)})}{|t^{(n)}|} = 0 \,\, .\] By claim \(iii)\) in Theorem 1, it follows that \({\rm span}_{\mathbb{R}}\{V_i, V_j\}\) is an abelian Lie subalgebra of \(\mathfrak{m}_0\), which contradicts the hypothesis \(\mathfrak{m}_0 \simeq \mathfrak{so}(3)\).

As a consequence, up to a permutation of the indices \(\{1,2,3\}\), it follows that \(\frac{1}{|t|}x_1(t) \to 0\) for \(t \to -\infty\). By applying claim \(iii)\) in Theorem 1 as above, since the rank of \(\mathfrak{m}_0\) is 1, it follows that \(\tfrac1{|t|}x_2(t)\) and \(\tfrac1{|t|}x_3(t)\) are bounded away from zero as \(t \to -\infty\). Hence, the collapsing torus is unique, and its Lie algebra coincides with \(\mathfrak{m}_1\). ◻

Proof. An explicit computation shows that \[\begin{align} &{\rm ric}_{\phi_{p}(i)}(g) -{\rm ric}_{i}(g) = \\ &\quad \overset{\eqref{eq:ric}}{=} \frac{b_{\phi_p(i)}}{2x_{\phi_p(i)}} -\frac{b_i}{2x_i} -\frac{1}{2d_{\phi_p(i)}}\sum_{1 \leq j,k \leq \ell}[\phi_p(i)jk]\frac{x_k}{x_{\phi_p(i)} x_j} +\frac{1}{2d_i}\sum_{1 \leq j,k \leq \ell}[ijk]\frac{x_k}{x_i x_j} \\ &\quad\qquad +\frac{1}{4d_{\phi_p(i)}}\sum_{1 \leq j, k \leq \ell}[\phi_p(i)jk]\frac{x_{\phi_p(i)}}{x_j x_k} -\frac{1}{4d_i}\sum_{1 \leq j, k \leq \ell}[ijk]\frac{x_i}{x_j x_k} \\ &\quad \overset{\eqref{eq:coef1}}{=} -\frac{b_i}{2}\bigg(\frac{1}{x_i} -\frac{1}{x_{\phi_p(i)}}\bigg) \\ &\quad\qquad -\frac{1}{2d_i}\sum_{1 \leq q \leq s}[q\phi_p(i)\phi_q(\phi_p(i))]\frac{x_{\phi_q(\phi_p(i))}}{x_{\phi_p(i)} x_q} +\frac{1}{2d_i}\sum_{1 \leq q \leq s}[iq\phi_q(i)]\frac{x_{\phi_q(i)}}{x_i x_q} \\ &\quad\qquad +\frac{1}{4d_i}\sum_{1 \leq q \leq s}[\phi_p(i)q\phi_q(\phi_p(i))]\frac{x_{\phi_p(i)}}{x_q x_{\phi_q(\phi_p(i))}} -\frac{1}{4d_i}\sum_{1 \leq q \leq s}[iq\phi_q(i)]\frac{x_i}{x_q x_{\phi_q(i)}} \\ &\quad\qquad -\frac{1}{2d_i}\sum_{1 \leq q \leq s}[\phi_p(i)\phi_q(\phi_p(i))q]\frac{x_q}{x_{\phi_p(i)} x_{\phi_q(\phi_p(i))}} +\frac{1}{2d_i}\sum_{1 \leq q \leq s}[i\phi_q(i)q]\frac{x_q}{x_i x_{\phi_q(i)}} \\ &\quad\qquad +\frac{1}{4d_i}\sum_{1 \leq q \leq s}[\phi_p(i)\phi_q(\phi_p(i))q]\frac{x_{\phi_p(i)}}{x_{\phi_q(\phi_p(i))} x_q} -\frac{1}{4d_i}\sum_{1 \leq q \leq s}[i\phi_q(i)q]\frac{x_i}{x_{\phi_q(i)} x_q} \\ &\quad\qquad -\frac{1}{2d_i}\sum_{\substack{s < j, k \leq \ell \\ j \in I_p^0 ,\;k \in I_p^0}}[\phi_p(i)jk]\frac{x_k}{x_{\phi_p(i)} x_j} +\frac{1}{2d_i}\sum_{\substack{s < j, k \leq \ell \\ j \in I_p^0 ,\;k \in I_p^0}}[ijk]\frac{x_k}{x_i x_j} \\ &\quad\qquad -\frac{1}{2d_i}\sum_{\substack{s < j, k \leq \ell \\ j \in I_p^+ ,\;k \in I_p^0}}[\phi_p(i)\phi_p(j)k]\frac{x_k}{x_{\phi_p(i)} x_{\phi_p(j)}} +\frac{1}{2d_i}\sum_{\substack{s < j, k \leq \ell \\ j \in I_p^+ ,\;k \in I_p^0}}[ijk]\frac{x_k}{x_i x_j} \\ &\quad\qquad -\frac{1}{2d_i}\sum_{\substack{s < j, k \leq \ell \\ j \in I_p^0 ,\;k \in I_p^+}}[\phi_p(i)j\phi_p(k)]\frac{x_{\phi_p(k)}}{x_{\phi_p(i)} x_j} +\frac{1}{2d_i}\sum_{\substack{s < j, k \leq \ell \\ j \in I_p^0 ,\;k \in I_p^+}}[ijk]\frac{x_k}{x_i x_j} \\ &\quad\qquad -\frac{1}{2d_i}\sum_{\substack{s < j, k \leq \ell \\ j \in I_p^+ ,\;k \in I_p^+}}[\phi_p(i)\phi_p(j)k]\frac{x_k}{x_{\phi_p(i)} x_{\phi_p(j)}} +\frac{1}{2d_i}\sum_{\substack{s < j, k \leq \ell \\ j \in I_p^+ ,\;k \in I_p^+}}[ijk]\frac{x_k}{x_i x_j} \\ &\quad\qquad +\frac{1}{4d_i}\sum_{\substack{s < j, k \leq \ell \\ j \in I_p^0 ,\;k \in I_p^0}}[\phi_p(i)jk]\frac{x_{\phi_p(i)}}{x_j x_k} -\frac{1}{4d_i}\sum_{\substack{s < j, k \leq \ell \\ j \in I_p^0 ,\;k \in I_p^0}}[ijk]\frac{x_i}{x_j x_k} \\ &\quad\qquad +\frac{1}{4d_i}\sum_{\substack{s < j, k \leq \ell \\ j \in I_p^+ ,\;k \in I_p^0}}[\phi_p(i)\phi_p(j)k]\frac{x_{\phi_p(i)}}{x_{\phi_p(j)} x_k} -\frac{1}{4d_i}\sum_{\substack{s < j, k \leq \ell \\ j \in I_p^+ ,\;k \in I_p^0}}[ijk]\frac{x_i}{x_j x_k} \\ &\quad\qquad +\frac{1}{4d_i}\sum_{\substack{s < j, k \leq \ell \\ j \in I_p^0 ,\;k \in I_p^+}}[\phi_p(i)j\phi_p(k)]\frac{x_{\phi_p(i)}}{x_j x_{\phi_p(k)}} -\frac{1}{4d_i}\sum_{\substack{s < j, k \leq \ell \\ j \in I_p^0 ,\;k \in I_p^+}}[ijk]\frac{x_i}{x_j x_k} \\ &\quad\qquad +\frac{1}{4d_i}\sum_{\substack{s < j, k \leq \ell \\ j \in I_p^+ ,\;k \in I_p^+}}[\phi_p(i)\phi_p(j)k]\frac{x_{\phi_p(i)}}{x_{\phi_p(j)} x_k} -\frac{1}{4d_i}\sum_{\substack{s < j, k \leq \ell \\ j \in I_p^+ ,\;k \in I_p^+}}[ijk]\frac{x_i}{x_j x_k} \,\, . \end{align}\] Therefore, by using 35 , 36 , 39 , 40 and 41 , we obtain \[\begin{align} &{\rm ric}_{\phi_{p}(i)}(g) -{\rm ric}_{i}(g) = \\ &\qquad = -\frac{b_i}{2}\bigg(\frac{1}{x_i} -\frac{1}{x_{\phi_p(i)}}\bigg) \\ &\quad\qquad -\frac{1}{2d_i}\sum_{1 \leq q \leq s}[qi\phi_q(i)]\bigg(\frac{x_{\phi_q(\phi_p(i))}}{x_{\phi_p(i)} x_q} -\frac{x_{\phi_p(i)}}{x_q x_{\phi_q(\phi_p(i))}}\bigg) +\frac{1}{2d_i}\sum_{1 \leq q \leq s}[qi\phi_q(i)]\bigg(\frac{x_{\phi_q(i)}}{x_i x_q} -\frac{x_i}{x_{\phi_q(i)} x_q}\bigg) \\ &\quad\qquad -\frac{1}{2d_i}\sum_{1 \leq q \leq s}[qi\phi_q(i)]\bigg(\frac{x_q}{x_{\phi_p(i)} x_{\phi_q(\phi_p(i))}} -\frac{x_q}{x_i x_{\phi_q(i)}}\bigg) \\ &\quad\qquad -\frac{1}{2d_i}\sum_{\substack{s < j,k \leq \ell \\ j \in I_p^0 ,\;k \in I_p^0}}[ijk]\bigg(\frac{x_k}{x_{\phi_p(i)} x_j} -\frac{x_k}{x_i x_j}\bigg) +\frac{1}{4d_i}\sum_{\substack{s < j, k \leq \ell \\ j \in I_p^0 ,\;k \in I_p^0}}[ijk]\bigg(\frac{x_{\phi_p(i)}}{x_j x_k} -\frac{x_i}{x_j x_k}\bigg) \\ &\quad\qquad -\frac{1}{2d_i}\sum_{\substack{s < j, k \leq \ell \\ j \in I_p^+ ,\;k \in I_p^0}}[ijk]\bigg(\frac{x_k}{x_{\phi_p(i)} x_{\phi_p(j)}} -\frac{x_k}{x_i x_j}\bigg) +\frac{1}{4d_i}\sum_{\substack{s < j, k \leq \ell \\ j \in I_p^+ ,\;k \in I_p^0}}[ijk]\bigg(\frac{x_{\phi_p(i)}}{x_{\phi_p(j)} x_k} -\frac{x_i}{x_j x_k}\bigg) \\ &\quad\qquad -\frac{1}{2d_i}\sum_{\substack{s < j, k \leq \ell \\ j \in I_p^0 ,\;k \in I_p^+}}[ijk]\bigg(\frac{x_{\phi_p(k)}}{x_{\phi_p(i)} x_j} -\frac{x_k}{x_i x_j}\bigg) +\frac{1}{4d_i}\sum_{\substack{s < j, k \leq \ell \\ j \in I_p^0 ,\;k \in I_p^+}}[ijk]\bigg(\frac{x_{\phi_p(i)}}{x_j x_{\phi_p(k)}} -\frac{x_i}{x_j x_k}\bigg) \\ &\quad\qquad -\frac{1}{2d_i}\sum_{\substack{s < j, k \leq \ell \\ j \in I_p^+ ,\;k \in I_p^+}}[ijk]\bigg(\frac{x_k}{x_{\phi_p(i)} x_{\phi_p(j)}} -\frac{x_k}{x_i x_j}\bigg) +\frac{1}{4d_i}\sum_{\substack{s < j, k \leq \ell \\ j \in I_p^+ ,\;k \in I_p^+}}[ijk]\bigg(\frac{x_{\phi_p(i)}}{x_{\phi_p(j)} x_k} -\frac{x_i}{x_j x_k}\bigg) \\ &\qquad = -\frac{b_i}{2}\bigg(\frac{1}{x_i} -\frac{1}{x_{\phi_p(i)}}\bigg) +\mu_{p,i}^2\bigg(\frac{x_{\phi_p(i)}}{x_i x_p} -\frac{x_i}{x_{\phi_p(i)} x_p}\bigg) \\ &\quad\qquad +\frac{1}{2}\sum_{\substack{1 \leq q \leq k \\ q \neq p}}\frac{\mu_{q,i}^2}{x_q}\bigg(\frac{x_{\phi_q(i)}}{x_i} -\frac{x_{\phi_q(\phi_p(i))}}{x_{\phi_p(i)}} +\frac{x_{\phi_p(i)}}{x_{\phi_q(\phi_p(i))}} -\frac{x_i}{x_{\phi_q(i)}}\bigg) \\ &\quad\qquad -\frac{1}{2}\sum_{\substack{1 \leq q \leq s \\ q \neq p}}\mu_{q,i}^2\bigg(\frac{x_q}{x_{\phi_p(i)} x_{\phi_q(\phi_p(i))}} -\frac{x_q}{x_i x_{\phi_q(i)}}\bigg) \\ &\quad\qquad -\frac{1}{2d_i}\sum_{\substack{s < j, k \leq \ell \\ j \in I_p^0 ,\;k \in I_p^0}}[ijk]\bigg(\frac{x_k}{x_{\phi_p(i)} x_j} -\frac{x_k}{x_i x_j}\bigg) +\frac{1}{4d_i}\sum_{\substack{s < j, k \leq \ell \\ j \in I_p^0 ,\;k \in I_p^0}}[ijk]\bigg(\frac{x_{\phi_p(i)}}{x_j x_k} -\frac{x_i}{x_j x_k}\bigg) \\ &\quad\qquad -\frac{1}{2d_i}\sum_{\substack{s < j, k \leq \ell \\ j \in I_p^+ ,\;k \in I_p^0}}[ijk]\bigg(\frac{x_k}{x_{\phi_p(i)} x_{\phi_p(j)}} -\frac{x_k}{x_i x_j}\bigg) +\frac{1}{4d_i}\sum_{\substack{s < j, k \leq \ell \\ j \in I_p^+ ,\;k \in I_p^0}}[ijk]\bigg(\frac{x_{\phi_p(i)}}{x_{\phi_p(j)} x_k} -\frac{x_i}{x_j x_k}\bigg) \\ &\quad\qquad -\frac{1}{2d_i}\sum_{\substack{s < j, k \leq \ell \\ j \in I_p^0 ,\;k \in I_p^+}}[ijk]\bigg(\frac{x_{\phi_p(k)}}{x_{\phi_p(i)} x_j} -\frac{x_k}{x_i x_j}\bigg) +\frac{1}{4d_i}\sum_{\substack{s < j, k \leq \ell \\ j \in I_p^0 ,\;k \in I_p^+}}[ijk]\bigg(\frac{x_{\phi_p(i)}}{x_j x_{\phi_p(k)}} -\frac{x_i}{x_j x_k}\bigg) \\ &\quad\qquad -\frac{1}{2d_i}\sum_{\substack{s < j, k \leq \ell \\ j \in I_p^+ ,\;k \in I_p^+}}[ijk]\bigg(\frac{x_k}{x_{\phi_p(i)} x_{\phi_p(j)}} -\frac{x_k}{x_i x_j}\bigg) +\frac{1}{4d_i}\sum_{\substack{s < j, k \leq \ell \\ j \in I_p^+ ,\;k \in I_p^+}}[ijk]\bigg(\frac{x_{\phi_p(i)}}{x_{\phi_p(j)} x_k} -\frac{x_i}{x_j x_k}\bigg) \,\, , \end{align}\] which concludes the proof. ◻

Proof. From 16 and 56 , it follows that \[\label{eq:AGAG3} \lim_{\tau \to +\infty} \frac{x_{\phi_p(i)}(\tau)}{x_i(\tau)} = 1 \quad \text{ for any 1 \leq p \leq s, for any i \in I_p. }\tag{57}\] Therefore, ?? follows by 53 , 54 and 57 . ◻

Proof. Notice that, for any \(\tau \geq 1\), \[x_j(\tau)\big(1 -\varepsilon_{p,j}(\tau)\big) \leq x_{\tilde{j}}(\tau) \leq x_j(\tau)\big(1 +\varepsilon_{p,j}(\tau)\big)\] and so \[\frac{x_{\tilde{j}}(\tau)}{x_{\tilde{k}}(\tau) x_{\tilde{r}}(\tau)} \leq \frac{x_j(\tau)}{x_k(\tau) x_r(\tau)} \frac{1 +\varepsilon_{p,j}(\tau)}{\big(1 -\varepsilon_{p,k}(\tau)\big)\big(1 -\varepsilon_{p,r}(\tau)\big)} \,\, .\] By 66 , it follows that \[0 \leq \varepsilon_{p,j}(\tau) \leq \frac{1}{10} \quad \text{for any j \in I_p}\] and so we have \[\begin{align} \frac{1 +\varepsilon_{p,j}(\tau)}{(1 -\varepsilon_{p,k}(\tau))(1 -\varepsilon_{p,r}(\tau))} &\leq \big(1 +\varepsilon_{p,j}(\tau)\big)\big(1 +2\varepsilon_{p,r}(\tau)\big)\big(1 +2\varepsilon_{p,k}(\tau)\big) \\ &\leq \big(1 +3\varepsilon_{p,j}(\tau) +3\varepsilon_{p,k}(\tau)\big)\big(1 +2\varepsilon_{p,r}(\tau)\big) \\ &\leq 1 +6\big(\varepsilon_{p,j}(\tau) +\varepsilon_{p,k}(\tau) +\varepsilon_{p,r}(\tau)\big) \end{align}\] and then 68 follows. ◻

Proof. As we mentioned in Section 2.1, it follows by hypothesis that \({\rm scal}(g(\tau))>0\) for any \(\tau > 0\). Moreover, by [51] and the fact that \(\mathfrak{t}\) is an abelian subalgebra of \(\mathfrak{m}_0\), we have \[\label{eq:WZ} d_qb_q = \sum_{s < j, k \leq \ell}[qjk] \quad \text{for any 1 \leq q \leq s} \, .\tag{70}\] Therefore, by 13 and 70 , we have \[\begin{align} {\rm scal}(g(\tau)) &= \frac{1}{2}\sum_{1 \leq j \leq \ell}\frac{d_j b_j}{x_j(\tau)} -\frac{1}{4}\sum_{1 \leq j, k, r \leq \ell}[jkr] \frac{x_j(\tau)}{x_k(\tau) x_r(\tau)} \\ &= \frac{1}{2}\sum_{1 \leq q \leq s}\frac{d_qb_q}{x_q(\tau)} -\frac{1}{2}\sum_{\substack{1 \leq q \leq s \\ s < j, k \leq \ell}}[qjk]\frac{x_k(\tau)}{x_q(\tau) x_j(\tau)} -\frac{1}{4}\sum_{\substack{1 \leq q \leq s \\ s < j, k \leq \ell}}[qjk]\frac{x_q(\tau)}{x_j(\tau) x_k(\tau)} \\ &\quad +\frac{1}{2}\sum_{s < j \leq \ell}\frac{d_jb_j}{x_j(\tau)} -\frac{1}{4}\sum_{s < j, k, r \leq \ell}[jkr]\frac{x_j(\tau)}{x_k(\tau) x_r(\tau)} \\ &= \frac{1}{2}\sum_{\substack{1 \leq q \leq s \\ s < j, k \leq \ell}}[qjk]\bigg(\frac{1}{x_q(\tau)} -\frac{x_k(\tau)}{x_q(\tau) x_j(\tau)}\bigg) -\frac{1}{4}\sum_{\substack{1 \leq q \leq s \\ s < j, k \leq \ell}}[qjk]\frac{x_q(\tau)}{x_j(\tau) x_k(\tau)} \\ &\quad +\frac{1}{2}\sum_{s < j \leq \ell}\frac{d_jb_j}{x_j(\tau)} -\frac{1}{4}\sum_{s < j, k, r \leq \ell}[jkr]\frac{x_j(\tau)}{x_k(\tau) x_r(\tau)} \\ &= -\frac{1}{2}\sum_{\substack{1 \leq q \leq s \\ j \in I_q}}[qj\phi_q(j)]\frac{(x_{\phi_q(j)}(\tau) -x_j(\tau))^2}{x_q(\tau) x_j(\tau) x_{\phi_q(j)}(\tau)} -\frac{1}{4}\sum_{\substack{1 \leq q \leq s \\ j \in I^+_q}}[qj\phi_q(j)]\frac{x_q(\tau)}{x_j(\tau) x_{\phi_q(j)}(\tau)} \\ &\quad +\frac{1}{2}\sum_{s < j \leq \ell}\frac{d_jb_j}{x_j(\tau)} -\frac{1}{4}\sum_{s < j, k, r \leq \ell}[jkr]\frac{x_j(\tau)}{x_k(\tau) x_r(\tau)} \\ &\leq \frac{1}{2}\sum_{s < j \leq \ell}\frac{d_jb_j}{x_j(\tau)} -\frac{1}{4}\sum_{s < j, k, r \leq \ell}[jkr]\frac{x_j(\tau)}{x_k(\tau) x_r(\tau)} \,\, , \end{align}\] which in turn implies that \[\sum_{s < j, k, r \leq \ell}[jkr]\frac{x_j(\tau)}{x_k(\tau) x_r(\tau)} < 2\sum_{s < j \leq \ell}\frac{d_jb_j}{x_j(\tau)} \leq \frac{2b_{\mathsf{G}/\mathsf{H}}}{\underline{x}(\tau)} \,\, ,\] where \(b_{\mathsf{G}/\mathsf{H}}\) and \(\underline{x}(\tau)\) were defined in 62 and 65 , respectively. Therefore, 69 follows. ◻

Proof. Let us observe that, from 54 and 64 , it follows that for any toral index \(1 \leq q \leq s\) and for any \(j \in I_q\) \[\label{def:thetaepsilon} \theta_{q,j}(\tau) = \frac{1}{4} \frac{x_j(\tau)}{x_{\phi_q(j)}(\tau)}\,\varepsilon_{q,j}(\tau)^2 \,\, .\tag{71}\] By 35 , 53 , 64 , 66 and 71 , there exists \(c(m)>1\) such that \[\label{eq:normSepsilon} \frac{1}{\alpha_p c(m)}\sum_{j \in I_p} \varepsilon_{p,j}(\tau) \leq |S_p(\tau)|_{g(\tau)} \leq \alpha_p c(m)\sum_{j \in I_p} \varepsilon_{p,j}(\tau) \quad \text{ for any \tau \geq 1} \,\, .\tag{72}\] Therefore, ?? follows from 68 , 69 and 72 . ◻

Proof. If \(x_{\phi_{p}(\iota)}(\tau^*) = x_{\iota}(\tau^*)\), the statement is trivially true. Assume \(x_{\phi_{p}(\iota)}(\tau^*) > x_{\iota}(\tau^*)\) and notice that 60 implies that \[\label{eq:max-p-i} \frac{x_{\phi_p{(\iota)}}(\tau^*)}{x_i(\tau^*)} \geq \max_{\substack{1 \leq q \leq s \\ 1 \leq j \leq \ell}}\bigg\{\frac{x_{\phi_q(j)}(t^*)}{x_j(\tau^*)} \, , \,\, \frac{x_j(\tau^*)}{x_{\phi_q(j)}(\tau^*)}\bigg\} \,\, .\tag{76}\] Indeed, observe that \[\theta_{p,\iota}(\tau^*) = f\left(\frac{x_{\phi_p(\iota)}(\tau^*)}{x_{\iota}(\tau^*)}\right) \,\, , \quad \text{ where f(y) \mathrel{\vcenter{:}}= \frac{1}{4}\left( y + \frac{1}{y} - 2 \right)} \,\, ,\] so 76 follows from the fact that \(f\) is a non-decreasing function of \(y \in \mathbb{R}\) for \(y \ge 1\).

Fix \(1 \leq q \leq s\). Then, by 36 and 76 , we get \[\begin{align} 0 &\leq \frac{x_{\phi_p{(\iota)}}(\tau^*)}{x_{\iota}(\tau^*)} -\frac{x_{\phi_p(\phi_q(\iota))}(\tau^*)}{x_{\phi_q(\iota)}(\tau^*)} = \frac{x_{\phi_p{(\iota)}}(\tau^*)}{x_{\phi_q(\iota)}(\tau^*)}\bigg(\frac{x_{\phi_q(\iota)}(\tau^*)}{x_{\iota}(\tau^*)} -\frac{x_{\phi_q(\phi_p(\iota))}(\tau^*)}{x_{\phi_p{(\iota)}}(\tau^*)}\bigg) \,\, , \\ 0 &\leq \frac{x_{\phi_p{(\iota)}}(\tau^*)}{x_{\iota}(\tau^*)} -\frac{x_{\phi_p(\phi_q(\iota))}(\tau^*)}{x_{\phi_q(\iota)}(\tau^*)} = \frac{x_{\phi_p(\phi_q(\iota))}(\tau^*)}{x_{\iota}(\tau^*)}\bigg(\frac{x_{\phi_p{(\iota)}}(\tau^*)}{x_{\phi_q(\phi_p(\iota))}(\tau^*)} -\frac{x_{\iota}(\tau^*)}{x_{\phi_q(\iota)}(\tau^*)}\bigg) \,\, , \\ 0 &\leq \frac{x_{\phi_p{(\iota)}}(\tau^*)}{x_{\iota}(\tau^*)} -\frac{x_{\phi_q(\iota)}(\tau^*)}{x_{\phi_p(\phi_q(\iota))}(\tau^*)} = x_{\phi_p(\iota)}(\tau^*) x_{\phi_q(\iota)}(\tau^*) \bigg(\frac{1}{x_{\iota}(\tau^*) x_{\phi_q(\iota)}(\tau^*)} -\frac{1}{x_{\phi_p(\iota)}(\tau^*) x_{\phi_q(\phi_p(\iota))}(\tau^*)}\bigg) \,\, . \end{align}\] Therefore, 73 , 74 and 75 are proved. If \(x_{\phi_{p}(\iota)}(\tau^*) < x_{\iota}(\tau^*)\), the proof goes the same way. ◻

Proof. By ?? and 59 , it follows that \[\begin{align} & \left.\frac{{\rm d}}{{\rm d} \tau}\theta_{p,{\iota}(p,\tau^*)}(\tau)\right|_{\tau = \tau^*} = \\ &= \frac{1}{2} \bigg(\frac{x_{\phi_p(\iota)}(\tau^*)}{x_{\iota}(\tau^*)} -\frac{x_{\iota}(\tau^*)}{x_{\phi_p(\iota)}(\tau^*)}\bigg) \Bigg\{\mu_{p,\iota}^2\frac{x_{\phi_p(\iota)}(\tau^*) -x_{\iota}(\tau^*)}{x_p(\tau^*) x_{\iota}(\tau^*) x_{\phi_p(\iota)}(\tau^*)}\bigg(x_{\phi_p(\iota)}(\tau^*) +x_{\iota}(\tau^*) -\frac{b_{\iota}}{2\mu_{p,\iota}^2}x_p(\tau^*)\bigg) \\ &\qquad +\frac{1}{2}\sum_{\substack{1 \leq q \leq s \\ q \neq p}}\mu_{q,\iota}^2x_q(\tau^*) \bigg(\frac{1}{x_{\iota}(\tau^*) x_{\phi_q(\iota)}(\tau^*)} -\frac{1}{x_{\phi_p(\iota)}(\tau^*) x_{\phi_q(\phi_p(\iota))}(\tau^*)}\bigg) \\ &\qquad +\frac{1}{2}\sum_{\substack{1 \leq q \leq s \\ q \neq p}}\frac{\mu_{q,\iota}^2}{x_q(\tau^*)}\bigg(\frac{x_{\phi_q(\iota)}(\tau^*)}{x_{\iota}(\tau^*)} -\frac{x_{\phi_q(\phi_p(\iota))}(\tau^*)}{x_{\phi_p(\iota)}(\tau^*)} +\frac{x_{\phi_p(\iota)}(\tau^*)}{x_{\phi_q(\phi_p(\iota))}(\tau^*)} -\frac{x_{\iota}(\tau^*)}{x_{\phi_q(\iota)}(\tau^*)}\bigg) \\ &\qquad -\frac{1}{2d_{\iota}}\sum_{\substack{s < j, k \leq \ell \\ j \in I_p^0 ,\;k \in I_p^0}}[\iota jk]\bigg(\frac{x_k(\tau^*)}{x_{\phi_p(\iota)}(\tau^*) x_j(\tau^*)} -\frac{x_k(\tau^*)}{x_{\iota}(\tau^*) x_j(\tau^*)}\bigg) \\ &\qquad +\frac{1}{4d_{\iota}}\sum_{\substack{s < j, k \leq \ell \\ j \in I_p^0 ,\;k \in I_p^0}}[\iota jk]\bigg(\frac{x_{\phi_p(\iota)}(\tau^*)}{x_j(\tau^*) x_k(\tau^*)} -\frac{x_{\iota}(\tau^*)}{x_j(\tau^*) x_k(\tau^*)}\bigg) \\ &\qquad -\frac{1}{2d_{\iota}}\sum_{\substack{s < j, k \leq \ell \\ j \in I_p^0 ,\;k \in I_p^+}}[\iota jk]\bigg(\frac{x_{\phi_p(k)}(\tau^*)}{x_{\phi_p(\iota)}(\tau^*) x_j(\tau^*)} -\frac{x_k(\tau^*)}{x_{\iota}(\tau^*) x_j(\tau^*)}\bigg) \\ &\qquad +\frac{1}{4d_{\iota}}\sum_{\substack{s < j, k \leq \ell \\ j \in I_p^0 ,\;k \in I_p^+}}[\iota jk]\bigg(\frac{x_{\phi_p(\iota)}(\tau^*)}{x_j(\tau^*) x_{\phi_p(k)}(\tau^*)} -\frac{x_{\iota}(\tau^*)}{x_j(\tau^*) x_k(\tau^*)}\bigg) \\ &\qquad -\frac{1}{2d_{\iota}}\sum_{\substack{s < j, k \leq \ell \\ j \in I_p^+}}[\iota jk]\bigg(\frac{x_k(\tau^*)}{x_{\phi_p(\iota)}(\tau^*) x_{\phi_p(j)}(\tau^*)} -\frac{x_k(\tau^*)}{x_{\iota}(\tau^*) x_j(\tau^*)}\bigg) \\ &\qquad +\frac{1}{4d_{\iota}}\sum_{\substack{s < j, k \leq \ell \\ j \in I_p^+}}[\iota jk]\bigg(\frac{x_{\phi_p(\iota)}(\tau^*)}{x_{\phi_p(j)}(\tau^*) x_k(\tau^*)} -\frac{x_{\iota}(\tau^*)}{x_j(\tau^*) x_k(\tau^*)}\bigg) \Bigg\} \,\, . \end{align}\] Then, by applying 73 , 74 and 75 , we get \[\begin{align} &\left.\frac{{\rm d}}{{\rm d} \tau}\theta_{p,{\iota}(p,\tau^*)}(\tau)\right|_{\tau = \tau^*} \geq \\ &\geq \frac{1}{2} \mu_{p,\iota}^2 \frac{\big(x_{\phi_p(\iota)}(\tau^*) -x_{\iota}(\tau^*)\big)^2\big(x_{\phi_p(\iota)}(\tau^*) +x_{\iota}(\tau^*)\big)}{x_{\iota}(\tau^*)^2 x_{\phi_p(\iota)}(\tau^*)^2 x_p(\tau^*)}\bigg(x_{\phi_p(\iota)}(\tau^*) +x_{\iota}(\tau^*) -\frac{b_{\iota}}{2\mu_{p,{\iota}}^2}x_p(\tau^*)\bigg) \\ &\qquad -\frac{1}{4d_{\iota}}\bigg|\frac{x_{\phi_p(\iota)}(\tau^*)}{x_{\iota}(\tau^*)} -\frac{x_{\iota}(\tau^*)}{x_{\phi_p(\iota)}(\tau^*)}\bigg|\Bigg\{\sum_{\substack{s < j, k \leq \ell \\ j \in I_p^0 ,\;k \in I_p^0}}[\iota jk]\bigg|\frac{x_k(\tau^*)}{x_{\phi_p(\iota)}(\tau^*) x_j(\tau^*)} -\frac{x_k(\tau^*)}{x_{\iota}(\tau^*) x_j(\tau^*)}\bigg| \\ &\qquad\qquad +\sum_{\substack{s < j, k \leq \ell \\ j \in I_p^0 ,\;k \in I_p^0}}[\iota jk]\bigg|\frac{x_{\phi_p(\iota)}(\tau^*)}{x_j(\tau^*) x_k(\tau^*)} -\frac{x_{\iota}(\tau^*)}{x_j(\tau^*) x_k(\tau^*)}\bigg| \\ &\qquad\qquad +\sum_{\substack{s < j, k \leq \ell \\ j \in I_p^0 ,\;k \in I_p^+}}[\iota jk]\bigg|\frac{x_{\phi_p(k)}(\tau^*)}{x_{\phi_p(\iota)}(\tau^*) x_j(\tau^*)} -\frac{x_k(\tau^*)}{x_{\iota}(\tau^*) x_j(\tau^*)}\bigg| \\ &\qquad\qquad +\sum_{\substack{s < j, k \leq \ell \\ j \in I_p^0 ,\;k \in I_p^+}}[\iota jk]\bigg|\frac{x_{\phi_p(\iota)}(\tau^*)}{x_j(\tau^*) x_{\phi_p(k)}(\tau^*)} -\frac{x_{\iota}(\tau^*)}{x_j(\tau^*) x_k(\tau^*)}\bigg| \\ &\qquad\qquad +\sum_{\substack{s < j, k \leq \ell \\ j \in I_p^+}}[\iota jk]\bigg|\frac{x_k(\tau^*)}{x_{\phi_p(\iota)}(\tau^*) x_{\phi_p(j)}(\tau^*)} -\frac{x_k(\tau^*)}{x_{\iota}(\tau^*) x_j(\tau^*)}\bigg| \\ &\qquad\qquad +\sum_{\substack{s < j, k \leq \ell \\ j \in I_p^+}}[\iota jk]\bigg|\frac{x_{\phi_p(\iota)}(\tau^*)}{x_{\phi_p(j)}(\tau^*) x_k(\tau^*)} -\frac{x_{\iota}(\tau^*)}{x_j(\tau^*) x_k(\tau^*)}\bigg| \Bigg\} \,\, . \end{align}\] By 54 , 63 and ?? , we get \[\begin{gather} \left.\frac{{\rm d}}{{\rm d} \tau}\theta_{p,{\iota}(p,\tau^*)}(\tau)\right|_{\tau = \tau^*} \geq \frac{2}{\alpha_p} \frac{\theta_{p,\iota}(\tau^*)}{x_p(\tau^*)} \bigg(\frac{x_{\phi_p(\iota)}(\tau^*) +x_{\iota}(\tau^*)}{x_{\phi_p(\iota)}(\tau^*)}\bigg) \Bigg(\frac{x_{\phi_p(i)}(\tau^*) +x_{\iota}(\tau^*) -\frac{b_i}{2\mu_{p,\iota}^2}x_p(\tau^*)}{x_{\iota}(\tau^*)}\Bigg) \\ -\frac{1}{2d_{\iota}}\bigg|\frac{x_{\phi_p(\iota)}(\tau^*)}{x_{\iota}(\tau^*)} -\frac{x_{\iota}(\tau^*)}{x_{\phi_p(\iota)}(\tau^*)}\bigg|\Bigg\{ 6(\ell-k)^2\cdot12\,b_{\mathsf{G}/\mathsf{H}}\, \alpha_p \, c(m) \frac{|S_p(\tau^*)|_{g(\tau^*)}}{\underline{x}(\tau^*)} \Bigg\} \,\, . \end{gather}\] Notice that, by 66 , we obtain \[\label{eq:obvious} \frac{5}{3} < \frac{x_{\phi_p(\iota)}(\tau^*) +x_{\iota}(\tau^*)}{x_{\phi_p(\iota)}(\tau^*)} < \frac{20}{9}\tag{78}\] and that, by recalling definitions 62 , 63 , 65 and by applying 66 , 67 , we get \[\label{eq:positiveaddend} \frac{x_{\phi_p(\iota)}(\tau^*) +x_{\iota}(\tau^*) -\frac{b_{\iota}}{2\mu_{p,\iota}^2}x_p(\tau^*)}{x_{\iota}(\tau^*)} \geq \frac{3}{2} -\frac{b_{\mathsf{G}/\mathsf{H}}\alpha_p^2}{2}\frac{x_p(\tau^*)}{\underline{x}(\tau^*)} > 0 \,\, .\tag{79}\] Therefore, by 64 , 78 and 79 , we get \[\left.\frac{{\rm d}}{{\rm d} \tau}\theta_{p,{\iota}(p,\tau^*)}(\tau)\right|_{\tau = \tau^*} \geq \frac{5}{\alpha_p} \frac{\theta_{p,i}(\tau^*)}{x_p(\tau^*)} \left(1 -\frac{b_{\mathsf{G}/\mathsf{H}}\alpha_p^2}{3}\frac{x_p(\tau^*)}{\underline{x}(\tau^*)}\right) -\frac{80 (\ell-k)^2\,b_{\mathsf{G}/\mathsf{H}}\, \alpha_p \, c(m)}{d_{\iota}} \varepsilon_{p,\iota}(\tau^*) \frac{|S_p(\tau^*)|_{g(\tau^*)}}{\underline{x}(\tau^*)} \,\, .\] By applying 35 , 53 , 63 , 60 and 72 , we get \[\label{eq:final} \left.\frac{{\rm d}}{{\rm d} \tau}\theta_{p,{\iota}(p,\tau^*)}(\tau)\right|_{\tau = \tau^*} \geq \frac{5}{d_{\iota} m \alpha_p^3} \Bigg(1 -\frac{b_{\mathsf{G}/\mathsf{H}}\alpha_p^2\big(1 +48 (\ell-k)^2\,m\alpha_p^3 \, c(m)^2\big)}{3} \frac{x_p(\tau^*)}{\underline{x}(\tau^*)} \Bigg) \frac{\big(|S_p(\tau^*)|_{g(\tau^*)}\big)^2}{x_p(\tau^*)} \,\, .\tag{80}\] By 56 and 65 , we can chose now a time \(T_p>0\) such that \[1 -\frac{b_{\mathsf{G}/\mathsf{H}}\alpha_p^2\big(1 +48 (\ell-k)^2\,m\alpha_p^3 \, c(m)^2\big)}{3} \frac{x_p(\tau^*)}{\underline{x}(\tau^*)} \geq \frac{1}{5} \quad \text{for any \tau \geq T_p}\] and so, if we set \[\delta(m,p) \mathrel{\vcenter{:}}= \frac{1}{m^2 \alpha_p^3} \,\, ,\] 77 immediately follows from 80 . ◻

Proof of Theorem 2. Let \(g(t)\) be a collapsed ancient solution to the homogeneous Ricci flow on a compact manifold. By abuse of notation, we set \(\tau = -t\) and denote by \(g(\tau)\) the corresponding backward solution. Assume that \(g(1)\), and hence \(g(\tau)\) for any \(\tau > 0\), is diagonal with respect to an NR-decomposition \(\varphi\). Assume the collapsing torus \(\mathsf{T}\) of \(g(\tau)\) is unique. Then we observe that, since the solution is \(\varphi\)-diagonal, \(\mathsf{T}\) is rigid (see Remark 4). Therefore, both the assumptions [40D41] and [40R41] are satisfied.

Now, fix a toral index \(1 \leq p \leq s\) and recall that, by 53 and ?? , \[\label{eq:finalfinal} \big(|S_p(\tau)|_{g(\tau)}\big)^2 = \sum_{j \in I_p} [pj\phi_p(j)]\, \theta_{p,j}(\tau) \,\rightarrow\, 0 \quad \text{as \tau \to +\infty} \,\, .\tag{81}\] If there exists \(\tau^* \geq 1\) such that \(|S_p(\tau^*)|_{g(\tau^*)}=0\), then \(|S_p(\tau)|_{g(\tau)}=0\) for any \(\tau > 0\) since isometries are preserved under the Ricci flow. Assume then by contradiction that \(|S_p(\tau)|_{g(\tau)}>0\) for any \(\tau > 0\). Then, by 60 and 77 , it follows that the largest summand \(\theta_{p,\iota(p,\tau)}(\tau)\) is strictly increasing as \(\tau \to +\infty\), which contradicts 81 . ◻

Proof. Consider the space of submersion metrics on \(M\) that are \(\varphi\)-diagonal and possibly not positive-definite on the toral fibers, i.e., \[\EuScript{N} \mathrel{\vcenter{:}}= \big\{ g = y_1\,Q|_{\mathfrak{m}_1{\otimes}\mathfrak{m}_1} + {\dots} + y_s\,Q|_{\mathfrak{m}_s{\otimes}\mathfrak{m}_s} + x_1\,Q|_{\mathfrak{b}_1{\otimes}\mathfrak{b}_1} + {\dots} + x_r\,Q|_{\mathfrak{b}_r{\otimes}\mathfrak{b}_r} : y_i \in \mathbb{R} \, , \,\, x_j>0 \big\} \,\, .\] By [34], the Ricci curvature can be extended analytically to the space \(\EuScript{N}\). Moreover, since \(\varphi\) is an NR-decomposition, the Ricci curvature is diagonal, so we obtain an analytic function \({\rm Ric} : \EuScript{N} \to \mathbb{R}^{s+r}\). Now let \[\check{g} = \check{x}_1\,Q|_{\mathfrak{b}_1{\otimes}\mathfrak{b}_1} + {\dots} + \check{x}_r\,Q|_{\mathfrak{b}_r{\otimes}\mathfrak{b}_r}\] be a \(\mathsf{G}\)-invariant, unit volume, \(\check{\varphi}\)-diagonal Einstein metric on \(B = \mathsf{G}/\mathsf{H}\mathsf{T}\) and consider the Euclidean scalar product on \(\mathbb{R}^{s+r}\) defined by \[\label{def:prod} \langle\!\langle (y_1,{\dots},y_s,x_1,{\dots},x_r)^t, (y'_1,{\dots},y'_s,x'_1,{\dots},x'_r)^t\rangle\!\rangle^{\mathsmaller{(\check{g})}} \mathrel{\vcenter{:}}= \sum_{q=1}^s \frac{1}{\dim(\mathfrak{b})}y_qy'_q + \sum_{k=1}^r \frac{\dim(\mathfrak{b}_k)}{\dim(\mathfrak{b})}\frac{x_k x'_k}{\check{x}_k^2} \,\, .\tag{82}\] We define then the \(\check{g}\)-projected Ricci curvature to be the tensor \[\label{def:R} R^{\mathsmaller{(\check{g})}}: \EuScript{N} \to \mathbb{R}^{s+r} \,\, , \quad R^{\mathsmaller{(\check{g})}}(g) \mathrel{\vcenter{:}}= {\rm Ric}(g) -\frac{\langle\!\langle {\rm Ric}(g), g\rangle\!\rangle^{\mathsmaller{(\check{g})}}}{\langle\!\langle g, g\rangle\!\rangle^{\mathsmaller{(\check{g})}}}g\tag{83}\] obtained by projecting the Ricci curvature to the unit sphere \[\Sigma^{\mathsmaller{(\check{g})}} \mathrel{\vcenter{:}}= \{g \in \EuScript{N} : \langle\!\langle g,g \rangle\!\rangle^{\mathsmaller{(\check{g})}} = 1\} \,\, .\] Therefore, by applying the Center Manifold Theorem [52] as in the proof of [34], it is possible to find an ancient solution to the \(\check{g}\)-projected Ricci flow \[\partial_t h(t) = -2R^{\mathsmaller{(\check{g})}}(h(t))\] that lives in \(\Sigma^{\mathsmaller{(\check{g})}} \cap (\mathbb{R}_{>0})^{s+r}\) and converges to \(0 \oplus \check{g}\) as \(t \to -\infty\). Notice that the \(\check{g}\)-projected Ricci flow is equivalent to the Ricci flow, and thus it yields a \(\varphi\)-diagonal Ricci flow solution \(g(t)\) on \(M\). Arguing as in the proof of [34], \(g(t)\) is ancient as well. Moreover, the corresponding curvature-normalized solution \(\bar{g}(t)\) converges, up to scaling, to \(\check{g}\) in the Gromov-Hausdorff topology as \(t \to -\infty\), and so \(g(t)\) is collapsed. Finally, by construction, \(\mathsf{T}\) is the collapsing torus of \(g(t)\) and is rigid. ◻

Proof. Let \(\varphi = (\mathfrak{m}_1, {\dots}, \mathfrak{m}_{\ell})\) and \(\EuScript{B}\) as in the hypotheses. Notice that, by Theorem 4, the decomposition \(\varphi\) is necessarily stably Ricci-diagonal. We must prove it is also normalizer-adapted (see Definition 3). Fix \(1 \leq p \leq \ell_0\), \(1 \leq i \leq \ell\) and assume that \([\mathfrak{m}_p, \mathfrak{m}_i] \neq \{0\}\). Since the action of \({\rm ad}(V_p)\) is by intertwining maps, the image \([\mathfrak{m}_p, \mathfrak{m}_i]\) is an irreducible \({\rm Ad}(\mathsf{H})\)-module equivalent to \(\mathfrak{m}_i\). Moreover, since \([\mathfrak{m}_p, \mathfrak{m}_i] \neq \{0\}\), there exist \(e_1 \in \EuScript{B} \cap \mathfrak{m}_i\) and \(e_2 \in \EuScript{B} \cap \mathfrak{m}_j\) such that \(Q([V_p,e_1],e_2) \neq 0\), for some \(\mathfrak{m}_j \simeq \mathfrak{m}_i\). By 84 , for any other \(\mathfrak{m}_{j'} \simeq \mathfrak{m}_i\), it follows that \(Q([V_p,e_1], w) = 0\) for any \(w \in \EuScript{B} \cap \mathfrak{m}_{j'}\), and so we get \([\mathfrak{m}_p, \mathfrak{m}_i] = \mathfrak{m}_j\). ◻

Proof. Notice that, by Schur’s Lemma, \(\mathfrak{m}_0\) is always \(g(t)\)-orthogonal to its \(Q\)-orthogonal complement \(\mathfrak{m}_0^{\perp}\) inside \(\mathfrak{m}\) for any \(t \leq 0\), i.e., the metric \(g(t)\) splits as \[g(t) = g_0(t) + g_0^{\perp}(t) \quad \text{ for any t \leq 0} \,\, ,\] with \[g_0(t) \mathrel{\vcenter{:}}= g(t)|_{\mathfrak{m}_0 \otimes \mathfrak{m}_0} \in S^2_+(\mathfrak{m}_0^*) \quad \text{and} \quad g_0^{\perp}(t) \mathrel{\vcenter{:}}= g(t)|_{\mathfrak{m}_0^{\perp} \otimes \mathfrak{m}_0^{\perp}} \in S^2_+((\mathfrak{m}_0^{\perp})^*)^{{\rm Ad}(\mathsf{H})} \,\, .\] Then, the identity component of the gauge group \(N_{\mathsf{G}}(\mathsf{H})/\mathsf{H}\) is isomorphic to \(\mathsf{SO}(3)\) or \(\mathsf{Sp}(1)\) and, by Remark 1, it acts transitively on the space of \(Q\)-orthonormal bases for \(\mathfrak{m}_0\). Thus, up to isometry, we may assume that \(g_0(-1)\) is diagonal with respect to the standard basis \(\EuScript{B}_{\rm st}\) of \(\mathfrak{m}_0\), that is \(\{E_{12}, E_{23}, E_{13}\}\) for \(\mathfrak{so}(3)\), or \(\{\mathtt{i}, \mathtt{j}, \mathtt{k}\}\) for \(\mathfrak{sp}(1)\).

Suppose \((N_{\mathsf{G}}(\mathsf{H})/\mathsf{H})_0 \simeq \mathsf{Sp}(1)\). The imaginary unit quaternions \(\mathtt{i}, \mathtt{j}, \mathtt{k} \in \mathsf{Sp}(1)\) give rise, via the conjugation action, to diffeomorphisms \(\Psi_{\mathtt{i}}, \Psi_{\mathtt{j}}, \Psi_{\mathtt{k}}: M \to M\), with derivatives \(\psi_{\mathtt{i}}, \psi_{\mathtt{j}}, \psi_{\mathtt{k}} : \mathfrak{m} \to \mathfrak{m}\), whose restrictions to \(\mathfrak{m}_0\) switch the signs two at a time, i.e., \[\begin{array}{ccc} \psi_{\mathtt{i}}(\mathtt{i}) = +\mathtt{i} \, , & \psi_{\mathtt{i}}(\mathtt{j}) = -\mathtt{j} \, , & \psi_{\mathtt{i}}(\mathtt{k}) = -\mathtt{k} \,\, , \\ \psi_{\mathtt{j}}(\mathtt{i}) = -\mathtt{i} \, , & \psi_{\mathtt{j}}(\mathtt{j}) = +\mathtt{j} \, , & \psi_{\mathtt{j}}(\mathtt{k}) = -\mathtt{k} \,\, , \\ \psi_{\mathtt{k}}(\mathtt{i}) = -\mathtt{i} \, , & \psi_{\mathtt{k}}(\mathtt{j}) = -\mathtt{j} \, , & \psi_{\mathtt{k}}(\mathtt{k}) = +\mathtt{k} \,\, . \end{array}\] Assume \(\mathfrak{m}_0^{\perp}\) does not contain any pair of equivalent \({\rm Ad}(\mathsf{H})\)-modules. By Remark 1, the linear maps \(\psi_{\mathtt{i}}, \psi_{\mathtt{j}}, \psi_{\mathtt{k}}\) preserve the \({\rm Ad}(\mathsf{H})\)-invariant submodules of \(\mathfrak{m}_0^{\perp}\). As a consequence, the diffeomorphisms \(\Psi_{\mathtt{i}}, \Psi_{\mathtt{j}}, \Psi_{\mathtt{k}}\) will in fact be isometries if \(g_0(t)\) is diagonal in the basis \(\EuScript{B}_{\rm st}\). Since isometries are preserved under the flow, \(\Psi_{\mathtt{i}}, \Psi_{\mathtt{j}}, \Psi_{\mathtt{k}}\) are isometries for \(g(t)\) at any \(t < 0\). Finally, a metric for which these maps are isometries must be diagonal in the basis \(\EuScript{B}_{\rm st}\). This shows that \(g(t)\) remains diagonal along the flow.

If \((N_{\mathsf{G}}(\mathsf{H})/\mathsf{H})_0 \simeq \mathsf{SO}(3)\), the same argument goes through with the discrete isometries coming from the elements \({\rm diag}(1,-1,-1), {\rm diag}(-1,1,-1), {\rm diag}(-1,-1,1) \in \mathsf{SO}(3)\). ◻

Proof. We can assume \(\mathfrak{m}_0 \simeq \mathfrak{so}(3)\), and the case \(\mathfrak{m}_0 \simeq \mathbb{R}\) can be treated similarly. By assumption, there is a unique \(\alpha_0 \in \mathcal{R}^+_{\mathfrak{m}}\) such that \(\mathfrak{m}_0 = {\rm span}_{\mathbb{R}}\{H_{\alpha_0}\} + \mathfrak{r}_{\alpha_0}\). Then, the real Cartan-Weyl basis of \(\mathfrak{m}\) is \[\label{eq:CWbasis} \EuScript{B} = \{H_{\alpha_0}\} \cup \{X_{\alpha}, Y_{\alpha} : \alpha \in \mathcal{R}^+_{\mathfrak{m}}\} \,\, .\tag{85}\] By hypothesis, there exists a decomposition \(\varphi = (\mathfrak{m}_1, \dots, \mathfrak{m}_\ell)\) such that each non-trivial \(\mathfrak{m}_i \in \varphi\) is a sum of real root spaces of \(\mathfrak{g}\). Then, one can assume that \[\mathfrak{m}_1 = {\rm span}_{\mathbb{R}}\{H_{\alpha_0}\} \,\, , \quad \mathfrak{m}_2 = {\rm span}_{\mathbb{R}}\{X_{\alpha_0}\} \,\, , \quad \mathfrak{m}_3 = {\rm span}_{\mathbb{R}}\{Y_{\alpha_0}\}\] and therefore \(\EuScript{B}\) is \(\varphi\)-adapted. Fix \(2 \leq i \leq j \leq \ell\) and take two vectors \(e_1 \in \EuScript{B} \cap \mathfrak{m}_i\), \(e_2 \in \EuScript{B} \cap \mathfrak{m}_j\). By 85 , there exist \(\alpha, \beta \in \mathcal{R}^+_{\mathfrak{m}}\) such that \(e_1 \in \mathfrak{r}_{\alpha}\) and \(e_2 \in \mathfrak{r}_{\beta}\). By assumption, at most one of \(\alpha + \beta, \alpha - \beta, \beta - \alpha\) lies in \(\mathcal{R}^+_{\mathfrak{m}}\) and so, by 88 and 89 , either \([e_1, e_2] \in \mathfrak{h}\) or there exists \(e_3 \in \EuScript{B} \cap \mathfrak{m}_k\) such that \([e_1,e_2]_{\mathfrak{m}} = e_3\), for some \(1 \leq k \leq \ell\). In particular, for any \(1 \leq r, s \leq \ell\) with \(\mathfrak{m}_r \simeq \mathfrak{m}_s\) and \(r \neq s\), we get \[Q([e_1,e_2],v)\,Q([e_1,e_2],w) = 0 \quad \text{for any v \in \EuScript{B} \cap \mathfrak{m}_r, w \in \EuScript{B} \cap \mathfrak{m}_s} \,\, .\] Finally, by the properties of the root spaces, it follows that \([H_{\alpha_0},e_1] = \pm \alpha(H_{\alpha_0}) \tilde{e}_1\), where \(\tilde{e}_1\) is the unique vector contained in \((\EuScript{B} \cap \mathfrak{r}_{\alpha}) \setminus \{e_1\}\). Therefore, we obtain that for any \(1 \leq r, s \leq \ell\) with \(\mathfrak{m}_r \simeq \mathfrak{m}_s\) and \(r \neq s\), the following condition holds: \[Q([H_{\alpha_0},e_1],v)\,Q([H_{\alpha_0},e_1],w) = 0 \quad \text{for any v \in \EuScript{B} \cap \mathfrak{m}_r, w \in \EuScript{B} \cap \mathfrak{m}_s} \,\, .\] This concludes the proof. ◻

Proof. In both cases, by Remark 12, if \(\alpha, \beta \in \mathcal{R}^+\), then at most one of \(\alpha + \beta\), \(\alpha - \beta\), \(\beta - \alpha\) lies in \(\mathcal{R}^+\). Therefore, the result follows from Proposition 10. ◻

Proof. Denote by \(\mathcal{R}^+ = \mathcal{R}^+_{\mathfrak{h}} \cup \mathcal{R}^+_{\mathfrak{m}}\) the set of positive roots of \(\mathsf{G}/\mathsf{H}\) according to the splitting \(\mathfrak{g} = \mathfrak{h} \oplus \mathfrak{m}\), by \(\EuScript{B}_{\mathfrak{h}}\) the real Cartan-Weyl basis for the Lie algebra \(\mathfrak{h}\), by \(\EuScript{B}_{\mathfrak{m}}\) the real Cartan-Weyl basis for \(\mathfrak{m}\) and let \(\varphi = (\mathfrak{m}_1, \dots, \mathfrak{m}_\ell)\) be a decomposition with respect to which \(\EuScript{B}\) is \(\varphi\)-adapted. Then, \(\mathfrak{g} \oplus \mathfrak{g}\) splits as \[\label{eq:decGxG} \mathfrak{g} \oplus \mathfrak{g} = \Delta \mathfrak{h} + \tilde{\Delta}\mathfrak{h} + \sum_{i = 1}^{\ell}\mathfrak{m}_i^{(1)} + \sum_{j = 1}^{\ell}\mathfrak{m}_j^{(2)} \,\, ,\tag{86}\] with \[\begin{gather} \Delta \mathfrak{h} \mathrel{\vcenter{:}}= {\rm Lie}(\Delta \mathsf{H}) = \{(H,H) : H \in \mathfrak{h}\} \,\, , \quad \tilde{\Delta}\mathfrak{h} \mathrel{\vcenter{:}}= (\Delta \mathfrak{h})^{\perp} \cap (\mathfrak{h} \oplus \mathfrak{h}) = \{(H,-H) : H \in \mathfrak{h}\} \,\, , \\ \mathfrak{m}_i^{(1)} \mathrel{\vcenter{:}}= \{(X,0) : X \in \mathfrak{m}_i\} \,\, , \quad \mathfrak{m}_j^{(2)} \mathrel{\vcenter{:}}= \{(0,X) : X \in \mathfrak{m}_j\} \,\, . \end{gather}\] Notice that the \({\rm Ad}(\Delta\mathsf{H})\)-modules \(\mathfrak{m}_i^{(1)}\) and \(\mathfrak{m}_j^{(2)}\) are irreducible, while the \({\rm Ad}(\Delta\mathsf{H})\)-action on \(\tilde{\Delta}\mathfrak{h}\) is equivalent to the \({\rm Ad}(\mathsf{H})\)-action on \(\mathfrak{h}\). Let us consider now the following sets of vectors: \[\EuScript{B}_{\tilde{\Delta}\mathfrak{h}} \mathrel{\vcenter{:}}= \{(H,-H) : H \in \EuScript{B}_{\mathfrak{h}}\} \,\, , \quad \EuScript{B}_{\mathfrak{m}}^{(1)} \mathrel{\vcenter{:}}= \{(X,0) : X \in \EuScript{B}_{\mathfrak{m}}\} \,\, , \quad \EuScript{B}_{\mathfrak{m}}^{(2)} \mathrel{\vcenter{:}}= \{(0,X) : X \in \EuScript{B}_{\mathfrak{m}}\} \,\, .\] By hypothesis, there is no \({\rm Ad}(\Delta\mathsf{H})\)-module in \(\tilde{\Delta}\mathfrak{h}\) that is equivalent to some \(\mathfrak{m}_i^{(1)}\) or \(\mathfrak{m}_j^{(2)}\). Then, one can check that \(\EuScript{B} \mathrel{\vcenter{:}}= \EuScript{B}_{\tilde{\Delta}\mathfrak{h}} \cup \EuScript{B}_{\mathfrak{m}}^{(1)} \cup \EuScript{B}_{\mathfrak{m}}^{(2)}\) is a strongly nice basis for the isotropy representation of \(\mathsf{G} {\times} \mathsf{G} / \Delta \mathsf{H}\). ◻

Proof of claim \(P(1)\). In order to prove claim \(P(1)\), by 13 we have \[{\rm scal}\big(\bar{g}^{(n)}\big) \leq \frac{1}{4\bar{x}^{(n)}_1}\Bigg(2\,{\rm Tr}_Q(-\mathcal{B}_{\mathfrak{g}}) -\sum_{1 \leq i,j,k \leq \ell}[ijk]^{(n)}\frac{\bar{x}^{(n)}_1 \bar{x}^{(n)}_i}{\bar{x}^{(n)}_j \bar{x}^{(n)}_k} \Bigg) \,\, .\] Therefore, since the scalar curvature of \(\bar{g}^{(n)}\) is uniformly bounded (see 3 ) and \(\bar{x}^{(n)}_1 \to 0\) as established before, we have \[4\,\bar{x}^{(n)}_1{\rm scal}\big(\bar{g}^{(n)}\big) \to 0\] and so there exists \(C>0\) such that \[\label{eq:ijkest-CB0} \sum_{1 \leq i,j,k \leq \ell}[ijk]^{(n)}\frac{\bar{x}^{(n)}_i}{\bar{x}^{(n)}_j \bar{x}^{(n)}_k} \leq \frac{C}{\bar{x}^{(n)}_1} \quad \text{ for any n \in \mathbb{N}} \,\, .\tag{91}\] As a direct consequence of 91 : \[\label{eq:claim-subalg0} \text{[ijk]^{(\infty)}>0 \,\,with\,\, i, j \in J_1,\, k \in \{1,{\dots},\ell\} \quad \Longrightarrow \quad k \in J_1 .}\tag{92}\] Indeed, assume by contradiction that there exist \(i, j \in J_1\) and \(k \in J_{> 1}\) such that \([ijk]^{(\infty)}>0\). Then, by 91 , up to changing the constant, we obtain \[\frac{\bar{x}^{(n)}_k}{\bar{x}^{(n)}_i} \leq C \,\, ,\] which is not possible. Therefore, 92 implies that \(\mathfrak{a}_1^{(\infty)}\) is a subalgebra of \(\mathfrak{g}\).

Let \(\{e_{\alpha}^{(n)}\}\) be a \(\varphi^{(n)}\)-adapted, \(Q\)-orthonormal basis for \(\mathfrak{m}\) and consider the quantities \[{\rm sec}^{(n)}_{ij} \mathrel{\vcenter{:}}= \sum_{e_{\alpha}^{(n)} \in \mathfrak{m}_i^{(n)}} \sum_{e_{\beta}^{(n)} \in \mathfrak{m}_j^{(n)}} {\rm sec}_{\bar{g}^{(n)}}\big(e_{\alpha}^{(n)},e_{\beta}^{(n)}\big) \,\, ,\] where \({\rm sec}\) denotes the sectional curvature. By [31], they satisfy the following relations for any \(1 \leq i < j \leq \ell\): \[\begin{align} 4\,\bar{x}_i^{(n)}{\rm sec}^{(n)}_{ij} &= \sum_{1 \leq k \leq \ell} [ijk]^{(n)} \frac{\bar{x}_i^{(n)}}{\bar{x}_j^{(n)}}\frac{\bar{x}_i^{(n)}}{\bar{x}_k^{(n)}} +\sum_{1 \leq k \leq \ell} [ijk]^{(n)} \bigg(\frac{\bar{x}_j^{(n)}}{\bar{x}_k^{(n)}} -1\bigg)\bigg(1 -2\frac{\bar{x}_i^{(n)}}{\bar{x}_j^{(n)}} +3\frac{\bar{x}_k^{(n)}}{\bar{x}_j^{(n)}}\bigg) \,\, , \tag{93} \\ 4\,\bar{x}_i^{(n)}{\rm sec}^{(n)}_{ii} &= 4d_ic_i^{(n)} +[iii]^{(n)} +4\sum_{\substack{1 \leq k \leq \ell \\ k \neq i}}[iik]^{(n)} -3 \sum_{\substack{1 \leq k \leq \ell \\ k \neq i}}[iik]^{(n)}\frac{\bar{x}_k^{(n)}}{\bar{x}_i^{(n)}} \,\, , \tag{94} \end{align}\] where \(c_i^{(n)}\) are the diagonal components of the Casimir operator (see [51]). The rest of the proof is divided into three steps.

Since the sectional curvature of \(\bar{g}^{(n)}\) is uniformly bounded, we have \[\lim_{n \to +\infty} \big(4\,\bar{x}_p^{(n)}{\rm sec}^{(n)}_{pi}\big) = 0 \quad \text{for any p \in J_1, i \in J_{>1} .}\] Noticing that \[\lim_{n \to +\infty} [pik]^{(n)} \frac{\bar{x}_p^{(n)}}{\bar{x}_i^{(n)}}\frac{\bar{x}_p^{(n)}}{\bar{x}_k^{(n)}} = 0 \quad \text{for any p \in J_1, i \in J_{>1}, k \in \{1,{\dots},\ell\} ,}\] equation 93 implies that \[\label{eq:proofAGAG0139} \lim_{n \to +\infty}\Bigg\{\sum_{1 \leq k \leq \ell} [pik]^{(n)} \bigg(\frac{\bar{x}_i^{(n)}}{\bar{x}_k^{(n)}} -1\bigg)\bigg(1 -2\frac{\bar{x}_p^{(n)}}{\bar{x}_i^{(n)}} +3\frac{\bar{x}_k^{(n)}}{\bar{x}_i^{(n)}}\bigg)\Bigg\} = 0 \quad \text{for any p \in J_1, i \in J_{>1} .}\tag{95}\] Arguing by induction as in [31], we obtain that each summand in 95 is infinitesimal, i.e., \[\label{eq:proofAGAG01} \lim_{n \to +\infty}\, [pik]^{(n)} \bigg(\frac{\bar{x}_i^{(n)}}{\bar{x}_k^{(n)}} -1\bigg)\bigg(1 -2\frac{\bar{x}_p^{(n)}}{\bar{x}_i^{(n)}} +3\frac{\bar{x}_k^{(n)}}{\bar{x}_i^{(n)}}\bigg) = 0 \quad \text{for any p \in J_1\,,\,\,i \in J_{>1}\,,\,\,k \in \{1,{\dots},\ell\}\,,}\tag{96}\] and so the following claims hold: \[\begin{align} \text{p \in J_1\,,\,\,\,i \in J_{>1}\,,\,\,\,k \in \{1,{\dots},\ell\}\,,\,\,\,[pik]^{(\infty)} = 0} \quad&\Longrightarrow\quad \lim_{n \to +\infty} [pik]^{(n)} \frac{\bar{x}_i^{(n)}}{\bar{x}_k^{(n)}} = 0 \,\, , \tag{97} \\ \text{p \in J_1\,,\,\,\,i, j \in J_{>1}\,,\,\,\,[pij]^{(\infty)} > 0} \quad&\Longrightarrow\quad \lim_{n \to +\infty} \frac{\bar{x}_i^{(n)}}{\bar{x}_j^{(n)}} = 1 \,\, . \tag{98} \end{align}\]

Since the sectional curvature of \(\bar{g}^{(n)}\) is uniformly bounded, we have \[\lim_{n \to +\infty} \big(4\,\bar{x}_p^{(n)}{\rm sec}^{(n)}_{pq}\big) = 0 \quad \text{for any p, q \in J_1\,,\,\,p \neq q\,.}\] Using 97 , we have \[\label{eq:proofAGAG2139} \lim_{n \to +\infty}\, [pqi]^{(n)} \bigg(\frac{\bar{x}_q^{(n)}}{\bar{x}_i^{(n)}} -1\bigg)\bigg(1 -2\frac{\bar{x}_p^{(n)}}{\bar{x}_q^{(n)}} +3\frac{\bar{x}_i^{(n)}}{\bar{x}_q^{(n)}}\bigg) = 0 \quad \text{for any p, q \in J_1\,,\,\,\,i \in J_{>1}\,.}\tag{99}\] Moreover \[\label{eq:proofAGAG213939} \lim_{n \to +\infty}\, [pqi]^{(n)} \frac{\bar{x}_p^{(n)}}{\bar{x}_q^{(n)}}\frac{\bar{x}_p^{(n)}}{\bar{x}_i^{(n)}} = 0 \quad \text{for any p, q \in J_1\,,\,\,\,i \in J_{>1}\,.}\tag{100}\] Therefore, 93 , 99 and 100 imply that \[\label{eq:proofAGAG21393939} \lim_{n \to +\infty} \Bigg\{\sum_{q' \in J_1} [pqq']^{(n)} \Bigg( \frac{\bar{x}_p^{(n)}}{\bar{x}_q^{(n)}}\frac{\bar{x}_p^{(n)}}{\bar{x}_{q'}^{(n)}} +\bigg(\frac{\bar{x}_q^{(n)}}{\bar{x}_{q'}^{(n)}} -1\bigg)\bigg(1 -2\frac{\bar{x}_p^{(n)}}{\bar{x}_q^{(n)}} +3\frac{\bar{x}_{q'}^{(n)}}{\bar{x}_q^{(n)}}\bigg) \Bigg) \Bigg\} = 0 \quad \text{for any p, q \in J_1\,,\,\,p \neq q\,.}\tag{101}\] Arguing by induction as in [31], we obtain that each summand in 101 is infinitesimal, i.e., \[\lim_{n \to +\infty} [pqq']^{(n)}\Bigg( \frac{\bar{x}_p^{(n)}}{\bar{x}_q^{(n)}}\frac{\bar{x}_p^{(n)}}{\bar{x}_{q'}^{(n)}} +\bigg(\frac{\bar{x}_q^{(n)}}{\bar{x}_{q'}^{(n)}} -1\bigg)\bigg(1 -2\frac{\bar{x}_p^{(n)}}{\bar{x}_q^{(n)}} +3\frac{\bar{x}_{q'}^{(n)}}{\bar{x}_q^{(n)}}\bigg)\Bigg) = 0 \quad \text{for any p \in J_1\,,\,\,q,q' \in J_1\,,\,\,q,q'>p}\] and so the following claim holds: \[\label{eq:profAGAG21} \text{p, q, q' \in J_1\,,\,\,\,p \leq q < q'} \quad\Longrightarrow\quad \lim_{n \to +\infty} [pqq']^{(n)} \frac{\bar{x}_{q'}^{(n)}}{\bar{x}_q^{(n)}} = 0 \,\, .\tag{102}\]

Since the sectional curvature of \(\bar{g}^{(n)}\) is uniformly bounded, we have \[\lim_{n \to +\infty} \big(4\,\bar{x}_p^{(n)}{\rm sec}^{(n)}_{pp}\big) = 0 \quad \text{for any p \in J_1\,.}\] By 96 and 97 , we have \[\lim_{n \to +\infty} [ppi]^{(n)} \frac{\bar{x}_i^{(n)}}{\bar{x}_p^{(n)}} = 0 \quad \text{for any p\in J_1\,,\,\,\,i \in J_{>1}}\] and so equation 94 implies that \[\lim_{n \to +\infty}\Bigg\{\sum_{q \in J_1 \setminus \{p\}}[ppq]^{(n)}\bigg(3\frac{\bar{x}_q^{(n)}}{\bar{x}_p^{(n)}} -4\bigg)\Bigg\} = 4d_p\,c_p^{(\infty)} +[ppp]^{(\infty)} \,\, .\] By 102 , the limit on the left-hand side is non-positive, and so we obtain \[\label{eq:profAGAG31} \text{p, q \in J_1} \quad\Longrightarrow\quad c_p^{(\infty)} = 0\,,\,\,[ppq]^{(\infty)} = 0 \,\, .\tag{103}\] Therefore, claim \(P(1)\) follows from 97 , 98 , 102 , 103 . ◻